Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 2nd Lesson Solutions Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 2nd Lesson Solutions

Very Short Answer Questions

Question 1.
Define the term solution.
Answer:
Solution: Solution is a homogeneous mixture of two (or) more components whose composition may be varied with certain limits.

Question 2.
Define molarity. [T.S. Mar.17] [Mar. 11]
Answer:
Molarity: The number of moles of solute dissolved in one litre of solution is called molarity.
Molarity (M) = \(\frac{\text { Moles of solute }}{\text { Volume (in litres) }}\)
M = \(\frac{\text { Weight }}{\text { Gram mol wt }} \times \frac{1000}{V(\mathrm{ml} l)}\)
Units: moles/litre.

Question 3.
Define molality [A.P. Mar. 15] [May 11]
Answer:
Molality: The number of moles of solute present in one kilogram of solvent is called molality.
Molality (M) = \(\frac{\text { Moles of solute }}{\text { Mass of solvent (kgs) }}\)
m = \(\frac{\text { Weight }}{\text { Gram mol wt }} \times \frac{1000}{\mathrm{G}(\mathrm{gms})}\)

Question 4.
Give an example of a solid solution in which the solute is solid.
Answer:
An example of a solid solution in which the solute is solid is copper dissolved in gold.

Question 5.
Define mole fraction. . [T.S. Mar. 18] [Mar. 14]
Answer:
Mole fraction: The ratio of number of moles of one component of the solution to the total number of moles of all the components of the solution is called mole fraction.
Mole fraction of solute X_s = \(\frac{\mathbf{n}_{\mathrm{S}}}{\mathbf{n}_{0}+\mathbf{n}_{\mathrm{s}}}\)
Mole fraction of solvent X₀ = \(\frac{\mathbf{n}_{\mathrm{S}}}{\mathbf{n}_{0}+\mathbf{n}_{\mathrm{s}}}\)
[n_s =number of moles of solute]
[n₀ = number of moles of solvent]
It has no units.

Question 6.
Define mass percentage solution.
Answer:
The mass percentage of a component of a solution is defined as the
mass % = × 100

Question 7.
What is ppm of a solution ?
Answer:
ppm – parts per million : It is a convenient method of expressing concentration when a solute is present in trace quantities. Parts per million is defined as the
ppm = × 10⁶

Question 8.
What role do the molecular interactions play in a solution of alcohol and water ?
Answer:
In a solution of alcohol and water the molecular interactions present are hydrogen bondings. When these components are mixed then new hydrogen bonds develop between alcohol and water. The formed forces are weaker. Due to the decrease of magnitude of attractive forces, the solution shows positive deviation from Raoult’s law. This leads to increase of vapour pressure of solution and decrease in its boiling point.

Question 9.
State Raoult’s law. [A.P. & T.S. Mar. 18; 16; A.P. Mar. 17] [Mar. 14]
Answer:
Raoult’s law for volatile solute: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

Raoult’s law for non-volatile solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute.

Question 10.
State Henry’s law.
Answer:
At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid. .
(Or)
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
P = K_H × x K_H = Henry’s constant
P = Partial pressure
x = Mole fraction of gas

Question 11.
What is Ebullioscopic constant ?
Answer:
Ebullioscopic constant : The elevation of boiling point observed in one molal solution containing non-volatile solute is called Ebullioscopic constant (or) molal elevation constant.

Question 12.
What is Cryoscopic constant ?
Answer:
Cryoscopic constant : The depression in freezing point observed in one molal solution containing non-volatile solute is called cryoscopic constant (or) molal depression constant.

Question 13.
Define osmotic pressure. [A.P. Mar. 18; 16]
Answer:
Osmotic pressure : The pressure required to prevent the in flow of solvent molecules into the solution when these are. separated by semi-permeable membrane.

Question 14.
What are isotonic solutions ? [T.S. Mar. 19; 16; A.P. Mar. 17, 15]
Answer:
Isotonic solutions : Solutions having same osmotic pressure at a given temperature are called isotonic solutions.
e.g.: Blood is isotonic with 0.9% \(\left(\frac{\mathrm{W}}{\mathrm{V}}\right)\) NaCl [Saline]

Question 15.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water, (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol.
Answer:

1. Phenol is partially soluble in water.
2. Toluene is insoluble in water.
3. Formic acid is highly soluble in water.
4. Ethylene glycol is highly soluble in water.
5. Chloroform is insoluble in water.
6. Pentanol is partially soluble in water.

Question 16.
Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 gm of C₉H₈O₄ is dissolved in 450g of CH₃CN.
Answer:
Given
The Mass of aspirin = 6.5 gms
Mass of acetonitrile = 450 gms
Mass of solution = 6.5 + 450
= 456.5 gms
Mass % = \(\frac{6.5}{456.5}\) × 100 = 1.424 %

Question 17.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 ml of 0.15 M solution in methanol.
Answer:
Given
Molarity = 0.15 M
Volume V = 250 ml
Molecular weight of benzoic acid (C₆H₅COOH) = 122
Molarity (M) = \(\frac{\text { Weight }}{\text { GMw }} \times \frac{1000}{V(\mathrm{~m} l)}\)
0.15 = \(\frac{\mathrm{W}}{122} \times \frac{1000}{250}\)
W = \(\frac{122 \times 0.15}{4}\) = 4.575 gms

Question 18.
The depression in freezing point of water observed for the same amount of acetic acid, dichloro-acetic acid and trichloro acetic acid increases in the order given above. Explain briefly.
Answer:
Given acids are CH₃COOH, CHCl₂COOH, CCl₃ COOH.
The depression in freezing point of a solute in water depends upon the number of ions (or) particles in aqueous solution.
The given three acids are arranged in the order of their acidic strengths (increasing order)
CH₃COOH < CHCl₂COOH < CCl₃COOH
Due to the presence of three chlorine atoms CCl₃COOH is more acidic than CHCl₂ COOH and followed by CH₃COOH.
∴ The order of depression in freezing point is
CH₃COOH < CHCl₂COOH < CCl₃COOH

Question 19.
What is Van’t Hoffs factor ‘i’ and how is it related to ‘a’ in the case of a binary electrolyte (1:1)?
Answer:
Van’t Hoffs factor (i) : “It is defined as the ratio of the observed value of colligative property to the theoretical value of colligative property”.
i =
Solute dissociation or ionization process : If a solute on ionization gives ‘n’ ions and ‘a’ is degree of ionization at the given concentration, we will have [1 + (n – 1) α] particles.
(1 – α) ⇌ nα
Total 1 – α + nα = [1 + (n – 1) α]
∴ i = \(\frac{[1+(n-1) \alpha]}{1}\)
α = \(\frac{\mathrm{i}-1}{\mathrm{n}-1}\)
α_ionization = \(\frac{\mathrm{i}-1}{\mathrm{n}-1}\)
Solute association process :
If ‘n’A molecules combine to give A_n, we have
nA ⇌ A_n
If ‘α’ is degree of association at the given concentration.
1 – α = \(\frac{\alpha}{n}\)
= 1 – α + \(\frac{\alpha}{n}\)
i = \(\frac{1-\alpha+\frac{\alpha}{n}}{1}\)
(or) α = \(\frac{1-\mathrm{i}}{1-\frac{1}{n}}\)
α _dissociation = \(\frac{1-\mathrm{i}}{1-\frac{1}{n}}\)

Question 20.
What is relative lowering of vapour pressure ? [A.P. Mar. 19]
Answer:
Relating lowering of vapour pressure: The ratio of lowering of vapour pressure of a solution containing non-volatile solute to the vapour pressure of pure solvent is called relative lowering of vapour pressure.
R.L.V.P. = \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{0}}\)
P₀ – P_S = lowering of vapour pressure
P₀ = Vapour pressure of pure solvent

Question 21.
Calculate the mole fraction of H₂SO₄ in a solution containing 98% H₂SO₄ by mass. [A.P. Mar. 17]
Answer:
Given a Solution containing – 98% H₂SO₄ by mass.
It means 98 gms of H₂SO₄ and 2 gms of H₂O mixed to form a solution.

Short Answer Questions

Question 1.
How many types of solutions are formed? Give an example for each type of solution.
Answer:
Solutions are classified into three types on the basis of the solvent present in that solution.

Question 2.
Define mass percentage, volume percentage and mass to volume percentage solutions.
Answer:
i) Mass percentage : The mass percentage of a component of a solution is defined as the
mass % = × 10
ii) Volume percentage \(\left(\frac{\mathbf{v}}{\mathbf{V}}\right)\) : It is defined as the volume % of a component
= × 100
iii) Mass to volume percentage \(\left(\frac{\mathbf{w}}{\mathbf{V}}\right)\) : It is defined as the mass of solute dissolved in 100 ml of the solution.
It is commonly used in medicine and pharmacy.

Question 3.
Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504g mL^-1?
Answer:
Given 68% HNO₃ by mass that means
68gms mass of HNO₃ present in 100 gms of solution.
Molecular weight of HNO₃ = 63
Number of moles of HNO₃ = \(\frac{\text { weight }}{\text { GMW }}=\frac{68}{63}\) = 1.079
Given density of the solution = 1.504 gm/ml

Question 4.
A solution of glucose in water is labelled as 10% w/w. What would be the molarity of the solution?
Answer:
Given 10% \(\left(\frac{\mathrm{W}}{\mathrm{w}}\right)\) glucose in water solution
Weight of glucose = 10 gms
GMW of glucose (C₆H₁₂O₆) = 180
Weight of water = 100 – 10 = 90 gms.
Molarity (m) = \(\frac{\text { Weight }}{\text { GMW }} \times \frac{1000}{\text { W (gms) }}\)
= \(\frac{10}{180} \times \frac{1000}{90}\) = \(\frac{100}{18 \times 9}\)
= 0.617 moles / kg
Molarity = \(\frac{\text { Weight }}{\text { GMW }} \times \frac{1000}{V(\mathrm{~m} l)}\)
Weight of solution = 100 gms
Assuming that Density of solution = 1.2 gm/ml
Volume = \(\frac{100}{1.2}\) = 83.33 ml
Molarity (M) = \(\frac{10}{180} \times \frac{1000}{83.33}\) ml
= \(\frac{1000}{18 \times 83.33}\) = 0.67 M

Question 5.
A solution of sucrose in water is labelled as 20% w/w. What would be the mole fraction of each component in the solution ?
Answer:
20% \(\left(\frac{\mathrm{W}}{\mathrm{w}}\right)\) Sucrose in water solution means
20 gms of Sucrose and 80 gms of water.
Number of moles of sucrose (n_s) = \(\frac{20}{342}\) = 0.0584
Number of moles of water (n₀) = \(\frac{80}{18}\) = 4.444
Mole fraction of sucrose X_s = \(\frac{n_{s}}{n_{0}+n_{s}}\)
= \(\frac{0.0584}{4.444+0.0584}\) + \(\frac{0.0584}{4.50284}\) = 0.01296
Mole fraction of water X₀ = \(\frac{n_{s}}{n_{0}+n_{s}}\) = \(\frac{4.444}{4.50284}\) = 0.9869
(Or)
X_s + X₀ = 1
X₀ = 1 – X_s = 1 – 0.01296 = 0987

Question 6.
How many ml of 0.1 M HCl is required to react completely with 1.0 g mixture of Na₂CO₃ and NaHCO₃ containing equal-molar amounts of both?
Answer:
Given 1gm mixture of Na₂CO₃ and NaHCO₃
Let the mass of Na₂CO₃ = a gms
Mass of NaHCO₃ = (1 – a) gms
Number of moles of Na₂CO₃ = \(\frac{\mathrm{wt}}{\text { GMW }}=\frac{\mathrm{a}}{106}\)
Number of moles of NaHCO₃ = \(\frac{\mathrm{wt}}{\mathrm{GMW}}=\frac{1-\mathrm{a}}{84}\)
Given that the mixture contains Equi molar amounts of Na₂CO₃ and NaHCO₂
∴ \(\frac{a}{106}\) = \(\frac{1-a}{84}\)
84 a =106 – 106 a
190 a = 106
a = 0.558gms
∴ Weight of Na₂CO₃ = 0.558 gms
Weight of NaHCO₃ = 1 – 0.558 = 0.442gms
Na₂CO₃ + 2 HCl → 2 NaCl + H₂O + CO₂
106 gms – 73 gms
0.558 g – ? \(\frac{73 \times 0.558}{106}\) = 0.384 gms
NaHCO₃ + HCl → NaCl + H₂O + CO₂
84 gms — 36.5 gins
0.442gms — ?
= \(\frac{36.5 \times 0.442}{84}\) = 0.1928
∴ The weight of HC1 required = 0.384 + 0.192 = 0.576 gms
Molarity (M) = \(\frac{\mathrm{Wt}}{\text { GMWt }} \times \frac{1000}{\mathrm{~V}(\mathrm{~m} l)}\)
0.1 = \(\frac{0.576}{36.5} \times \frac{1000}{V}\)
V = \(\frac{0.576 \times 1000}{36.5 \times 0.1}=\frac{576}{3.65}\) = 157.80 ml

Question 7.
A solution is obtained by mixing 300g of 25% solution and 400g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer:
Given a solution is obtained by mixing 300 gms of 25% solution and 400 gms of 40% solution by mass.
Weight of solute in 1st solution = 300 × \(\frac{25}{100}\) = 75 gms
Weight of solute in 2nd solution = 400 × \(\frac{40}{100}\) = 160 gms
Total weight of solute = 75 + 160 = 235 gms
Total weight of solution = 300 + 400 = 700 gms
Mass % of solute in resulting solution = \(\frac{235}{700}\) × 100 = 33.5%

Question 8.
An antifreeze solution is prepared from 222.6g of ethylene glycol [(C₂H₆O₂)] and 200g of water (solvent). Calculate the molality of the solution.
Solution:
Weight of Ethylene glycol = 222.6 gms .
G.mol wt =62
Weight of solvent = 200 gms
Molality (m) = \(\frac{\mathrm{Wt}}{\mathrm{GMWt}} \times \frac{1000}{\mathrm{a}(\mathrm{gms})}\)
= \(\frac{222.6}{62} \times \frac{1000}{200}=\frac{222.6}{62 \times 0.2}\) = 17.95 m
Molarity (M) = \(\frac{\mathrm{Wt}}{\mathrm{GMWt}} \times \frac{1000}{\mathrm{~V}(\mathrm{~m} l)}\)
= \(\frac{222.6}{62} \times \frac{1000}{394.2}\) = 9.1 M
Mass of solution = 200 + 222.6
= 422.6 gms
Volume = \(\frac{\text { Weight }}{\text { Density }}\) = \(\frac{422.6}{1.072}\) = 394.2 ml

Question 9.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Gas dissolved in a liquid is an exothermic reaction (H <O).
According to Lechatelier’s principle,

• If the reaction is exothermic the solubility of gas, decrease with increases in temperature.
• So gases always tend to be less soluble in liquids to as the temperature is raised.

Question 10.
What is meant by positive deviations from Raoults law and how is the sign of ∆_mix H related to positive deviation from Raoult’s law?
Answer:

• When the vapour pressure of a solution is higher than the predicted value by Raoult’s law, it is called positive deviation.
  
• In this case inter molecular interactions between solute and solvent particles (1 and 2) are weaker than those between solute and solvent particles (1 and 2) are weaker than those between solute and solute (1 – 1) and solvent – solvent (2 – 2)
• Hence the molecules of 1 or 2 will escape more easily from the surface of solution than in their pure state.
  ∴ The vapour pressure of the solution will be higher.
• The vapour pressure diagram showing positive deviation as follows.
  Eg. : Ethyl alcohol and water, Acetone and benzene.

Question 11.
What is meant by negative deviation from Raoult’s law and how is the sign of ∆_mixH related to negative deviation from Raoult’s law?
Answer:

• When the vapour pressure of a solution is lower than the predicted value by Raoult’s law, it is called negative deviation.
  
• In case of negative deviation the inter molecular attractive forces between 1 – 1 and 2 – 2 are weaker than those between 1 -2.
• It leads to decrease in vapour pressure resulting in negative deviation.
• The vapour pressure diagram showing negative deviation as follows.
• Eg.: HNO₃ and water, HCl and water.

Question 12.
The vapour pressure of water is 12.3 k P_a at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
Given molality of solution = 1 m
Vapour pressure of water P₀ = 12.3 kP_a
Number of moles of water = \(\frac{100}{18}\)
= 55.55 (n₀)
Mole fraction of solute (X_s) = \(\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}\)
= \(\frac{1}{55.55+1}\)
= \(\frac{1}{56.55}\) = 0.0177
Mole fraction of water (X₀) = 1 – X_s
= 1 – 0.0177 = 0.9823
Vapour pressure of solution (P_s) = P₀X₀
= 12.3 × 0.9823 = 12.08 kP_a

Question 13.
Calculate the mass of a non-volatile solute (molar mass 40g mol^-1) which should be dissolved in 114g Octane to reduce its vapour pressure to 80%. [T.S. Mar. 16]
Answer:
Raoult’s law formula

Given vapour pressure reduced to 80%, when non-volatile solute is dissolved in octane.
P₀ = 1 atm
P_s = 0.8 atm
W = 114 gms
M = 114 gm/mole
w = ? ; m = 40
\(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}}=\frac{\mathrm{w}}{\mathrm{m}} \times \frac{\mathrm{M}}{\mathrm{W}}\)
\(\frac{1-0.8}{1}=\frac{w}{40} \times \frac{114}{114}\)
w = 8 gms

Question 14.
A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water If freezing point of water is 273.15 K.
Answer:
Given 5% solution of cane sugar
w = 5gms; W = 95gms
m = 342; ∆T_f = 273.15 – 271 = 2.15
∆T_f = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}}{\mathrm{m} \times \mathrm{W}}\)
2.15 =\(\frac{\mathrm{K}_{\mathrm{f}} \times 5}{342 \times 95}\) ……………… (1)
5% solution of glucose
w = 5gms; W = 95gms
m = 180, ∆T_f = ?
∆T_f = \(\frac{K_{f} \times 5}{180 \times 95}\) ………………… (2)
Dividing equation (2) by Equation (1)
\(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{2.15}=\frac{\mathrm{K}_{\mathrm{f}} \times 5}{180 \times 95} \times \frac{342 \times 95}{\mathrm{~K}_{\mathrm{f}} \times 5}\)
∆T_f = \(\frac{342 \times 2.15}{180}\) = 4.085 K
∴ The freezing point temperature for 5% glucose solution
= 273.15 – 4.085 = 269.07 K

Question 15.
If the osmotic pressure of glucose solution is 1.52 bar at 300 K. What would be its concentration if R = 0.083L bar mol^-1 K^-1 ?
Answer:
π = CRT
R = 0.0836.bar. mole^-1 K^-1
T = 300 K
π = 1.52 bar
C = \(\frac{\pi}{\mathrm{RT}}=\frac{1.52}{0.083 \times 300}=\frac{1.52}{24.9}\) = 0.061 M

Question 16.
Vapour pressure of of water at 293K is 17.535 mm Hg. Calculate the vapour pressure of the solution at 293K when 25g of glucose is dissolved in 450g of water? [A.P. Mar.19]
Answer:
Raoult1s law formula

Question 17.
How is molar mass related to the elevation in boiling point of a solution?
Answer:
The expression for elevation of boiling point is
∆T_b = \(\frac{\mathrm{K}_{\mathrm{b}} \times 1000 \times \mathrm{w}}{\mathrm{m} \times \mathrm{w}}\) = K_b = molal elevation constant
w = Weight of solute
W = Weight of solvent
m = molar mass of solute
Molar mass of solute m = \(\frac{K_{\mathrm{b}} \times 1000 \times \mathrm{w}}{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{w}}\)
∴ Molar mass of solute (m) and elevation of boiling point (∆T_b) are inversely related.

Question 18.
What is an ideal solution ?
Answer:
A solution of two or mote components which obeys Raoult’s law at all concentrations and at all temperatures is called ideal Solution. In ideal solution there should not be any association between solute and solvent, (i.e.) no chemical interaction between solute and solvent of solution.
Ex : The following mixtures form ideal solutions.

• Benzene + Toluene
• n – hexane + n – heptane
• ethyl bromide + ethyl iodide

Question 19
What is relative lowering of vapour pressure ? How is it useful to determine the molar mass of a solute ? [T.S. Mar. 15]
Answer:
Raoult’s law for volatile solute: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

Raoult’s law for non-volatile solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute.

Relative lowering of vapour pressure \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}\) = X_s (mole fraction of solute)
\(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}\)
For very much dilute solutions n_s < < < …………… n₀
∴ \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}}=\frac{\mathrm{w}}{\mathrm{m}} \times \frac{\mathrm{M}}{\mathrm{W}}\)
W = Weight of solute
w = Weight of solvent
m = Molar mass of solute
M = Molar mass of solvent
Molar mass of solute m = \(\frac{W \times M}{W} \times \frac{P_{0}}{P_{0}-P_{s}}\)

Question 20.
How is molar mass related to the depression in freezing point of a solution ?
Answer:
The expression for depression in Freezing point is
∆T_f = \(\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{w}}{\mathrm{m} \times \mathrm{W}}\)
K_f = molal depression constant
W = Weight of solvent
w = Weight of solute
m = molar mass of solute
Molar mass of solute (m) = \(\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{w}}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{W}}\)
∴ Molar mass ôf solute (m) and depression in freezing point (∆T_f) are inversely related.

Long Answer Questions

Question 1.
An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute ?
Answer:
Relative lowering of vapour pressure = \(\frac{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{0}}=\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}}\)
P₀ = 1.013 bar, P_s = 1.004 bar
w = 2gms
W = 98gms
M = 18
m = ?

Question 2.
Heptane and Octane form an ideal solution. At 373 K the vapour pressure of the two liquid components are 105.2 kP_a and 46.8 kP_a respectively. What will be the vapour pressure of a mixture of 26.0 g heptane and 35g of octane?
Answer:
Number of moles of octane n₀ = \(\frac{\text { Weight }}{\text { GMwt }}=\frac{35}{114}\) = 0.307
Number of moles of heptane n_s = \(\frac{\text { Weight }}{\text { GMwt }}=\frac{26}{100}\) = 0.26
no 0.307
Mole fraction of octane X₀ = \(\frac{\mathrm{n}_{0}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}=\frac{0.307}{0.307+0.26}\) = 0.541
Mole fraction of heptane X_s = \(\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_{0}+\mathrm{n}_{\mathrm{s}}}=\frac{0.26}{0.307+0.26}\) = 0.459
Given vapour pressure of heptane P₁ = 105.2 kP_a
vapour pressure of Octane P₂ = 46.8 kP_a
26 gms of heptane and 35 gms of octane are mixed.
In that mixture
The vapour pressure of heptane (P₁₁) = P₁ × X_s
= 105.2 × 0.459 = 48.28 kP_a
The vapour pressure of Octane (P₂₂) = P₂ × X₀
= 46.8 × 0.541
= 25.32 kP_a
Total pressure of the mixture (P) = P₁₁ + P₂₂
= 25.32 + 48.28
= 73.6 kP_a

Question 3.
A solution containing 30g of non-volatile solute exactly in 90g of water has a vapour pressure of 2.8 kP_a at 298 K. Further 18g of water is then added to the solution and the new vapour pressure becomes 2.9 kP_a at 298 K. Calculate (i) The molar mass of the solute and (ii) Vapour pressure of water at 298 K.
Answer:
Calculation of molar mass of solute
Case – I:

Case – II :

Calculation of vapour pressure of water :
According to Raoult’s law
P_A = \(\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}\)
2.8 = \(P_{A}^{0} \times \frac{M}{6+M}\)
M = 23
2.8 = \(\mathrm{P}_{\mathrm{A}}^{0} \times \frac{23}{6+23}\)
\(\mathrm{P}_{\mathrm{A}}^{0}=\frac{2.8 \times 29}{23}=\frac{81.2}{23}\) = 3.53 k.p_a

Question 4.
Two elements A and B from compounds having formula AB₂ and AB₄. When dissolved in 20g of Benzene (C₆H₆), 1g of AB₂ lowers the freezing point by 2.3 K whereas 1.0g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol_-1. Calculate atomic masses of A and B.
Answer:
Calculation of molecular masses of compounds AB₂ and AB₄
For AB₂ compounds
m = \(\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{W}}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{W}}=\frac{5.1 \times 1000 \times 1}{2.3 \times 20}\) = 110.87 gms/mole
For AB₄ compound
m = \(\frac{5.1 \times 1000 \times 1}{1: 3 \times 20}\) = 196.15 gms/mole
Calculation of the atomic masses of elements
Atomic mass of element A = x
Atomic mass of element B = y
Molecular mass of AB₂ = x + 2y
Molecular mass of AB₄ = x + 4y
a + 2b = 110.87 ……………….. (1)
a + 4b = 196.15 ………………. (2)
Equation (2) – Eqiation (1)
x + 4y – x – 2y
196.15 – 110.87
2y = 85.28
y = 42.64
x + 2y = 110.87
x + 85.28 = 110.87
x = 25.59
Atomic mass of element A = 25.59 u
B = 42.64 u

Question 5.
Calculate the depression in the freezing point of water when 10g of CH₃CH₂CHClCOOH is added to 250g of water. K_a = 1.4 × 10^-3, K_f = 1.86 K kg mol^-1.
Answer:
Calculation of degree of dissociation :
Mass of solute = 10 gms
Molar mass of solute (CH₃ – CH₂ – CH – Cl – COOH) = 122.5 g/mole
Molality = \(\frac{10}{122.5} \times \frac{1000}{250}\) = 0.326 m

Calculation of depression in freezing point
∆T_f = i × K_f × m = 1.065 × 1.86 × 0.326 = 0.65 K

Question 6.
19.5g of CH₂FCOOH is dissolved in 500g of water. The depression In freezing point of water observed is 1.0°C. Calculate the Van’t Hoff factor and dissociation constant of fluoroacetic acid.
Answer:
Calculation of Vant Hoff factor of acid:
∆T_f = 1°C
K_f = 1.86 K kg/mole^-1
∆T_f = i K_f m
m = \(\frac{19.5}{78} \times \frac{1000}{500}\) = 0.5 m
i = \(\frac{1}{1.86 \times 0.5}\) = 1.0753
Calculation of degree of dissociation of the acid :

Question 7.
100 g of liquid A(molar mass 140g mol^-1) was dissolved in 1000 g of liquid B(molar mass 180g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.
Answer:
Calculation of vapour pressure of pure liquid A (\(\mathrm{P}_{\mathrm{A}}^{0}\))
Number of moles of liquid A
n_A = \(\frac{\mathrm{w}}{\mathrm{m}}=\frac{100}{140}\) = 0.7143
Number of moles of liquid B
n_B = \(\frac{\mathrm{w}}{\mathrm{m}}=\frac{1000}{180}\) = 5.5556
Mole fraction of A
X_A = \(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}}}=\frac{0.7143}{0.7143+5.5556}=\frac{0.7143}{6.2699}\)
= 0.1139
Mole fraction of B (X_B) = 1 – 0.1139 = 0.8861
Vapour pressure of liquid (\(\mathrm{P}_{\mathrm{B}}^{0}\)) = 500 torr
Total vapour pressure of solution (P) = 475 torr
According to Raoult’s law
P = P\(\mathrm{P}_{\mathrm{A}}^{0}\) X_A + P\(\mathrm{P}_{\mathrm{B}}^{0}\) X_B
475 = P\(\mathrm{P}_{\mathrm{A}}^{0}\) (0.1139) + 500(0.8861)
475 = P\(\mathrm{P}_{\mathrm{A}}^{0}\) (0.1139) + 443.05
\(\mathrm{P}_{\mathrm{A}}^{0}\) = \(\frac{475-443.05}{0.1139}=\frac{31.95}{0.1139}\) = 280.5 torr
Calculation of vapour pressure of A in the solution (P_A)
P_A = P\(\mathrm{P}_{\mathrm{A}}^{0}\) X_A = 280.5 × 0.1139
P_A = 32 torr.

Question 8.
Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Answer:
Varit Hoffs equation
Osmotic pressure (π) = i CRT
i = 2.47 .
V = 2.5 lit
R = 0.0821 lit. atm.k^-1. mol^-1
T = 27 + 273 = 300 K
π = 0.75 atm
π = i(\(\frac{n_{B}}{V}\))RT
n_B = \(\frac{\pi \mathrm{V}}{\mathrm{iRT}}=\frac{0.75 \times 2.5}{2.47 \times 0.0821 \times 300}\) = 0.0308
The amount of CaCl₂ dissolved = n_B × m_B
= 0.0308 × 111 = 3.42 gms

Question 9.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in two litre of water at 25°C assuming that it is completely dissociated.
Answer:
The amount of K₂SO₄ dissolved = 25 mg
Volume = 2 lit; T = 25°C = 298 K
Molecular weight of K₂SO₄ = 174 gms / mole

Total ions produced on dissociation = 3
i = 3
π = i CRT = \(\mathrm{i}\left(\frac{\mathrm{n}}{\mathrm{V}}\right) \mathrm{RT}=\mathrm{i}\left(\frac{\mathrm{w}}{\mathrm{m}}\right) \times \frac{\mathrm{RT}}{\mathrm{V}}\)
= \(\frac{3 \times 0.025 \times 0.0821 \times 298}{174 \times 2}\) = 5.27 × 10^-3 atm.

Question 10.
Benzene and Toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm of Hg and 32.06 mm of Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.
Answer:
Molecular weight of benzene (C₆H₆) = 78
Molecular weight of Toluene (C₇H₈) = 92
n_C6H6 = \(\frac{80}{78}\) = 1.026
n_C7H8= \(\frac{100}{92}\) = 1.087
X_C6H6 = \(\frac{1.026}{1.026+1.087}=\frac{1.026}{2.113}\) = 0.4855
X_C7H8 = 1 – 0.4855 = 0.5145
According to Raoult’s law
Partial pressure of benzene in solution
P_C6H6 = P_C6H6 × X_C6H6
= 50.71 × 0.4855 = 24.61 mm
Partial pressure of Toluene in solution
P_GH8 = × X_C7H8 = 32.06 × 0.5145 = 16.49 mm
Total vapour pressure of solution (P) = 24.61 + 16.49 = 41.1 mm
Mole fraction of benzene in vapour phase is
= \(\frac{\mathrm{X}_{\mathrm{C}_{6} \mathrm{H}_{6}} \times \mathrm{P}_{\mathrm{C}_{6} \mathrm{H}_{6}}^{0}}{\mathrm{P}_{\text {total }}}=\frac{0.4855 \times 50.71}{41.1}=\frac{24.61}{41.1}\) = 0.5987

Textual Examples

Question 1.
Calculate the mole fraction of ethylene glycol (C₂H₆O₂) In a solution containing 20% of C₂H₆O₂ by mass.
Solution:
Assume that we have 100g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20g of ethylene glycol and 80g of water.
Molar mass of C₂H₆O₂ = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol^-1

Question 2.
Calculate the molarity of a solution containing 5g of NaOH in 500 mL solution. [T.S Mar.17]
Solution:
Moles of NaOH = \(\frac{5 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.125 mol
Volume of the solution in litres = 450 mL / 1000 mL L^-1
Using equation

= 0.278 mol L^-1 = 0.278 mol dm^-3

Question 3.
Calculate molality of 2.5 of ethanoic acid (CH₃COOH) In 75g of benzene. [T.S. Mar. 15]
Solution:
Molar mass of C₂H₄O₂: 12 × 2 + 1 × 4 + 16 × 2 = 60g mol^-1
Moles of C₂H₄O₂ = \(\frac{2.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.0417 mol
Mass of benzene in kg = 75g/1000 g kg^-1 = 75 × 10^-3 kg
Molality of C₂H₄O₂ = \(\frac{\text { Moles of } \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}}{\mathrm{~kg} \text { of benzene }}=\frac{0.0417 \times \mathrm{mol} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{75 \mathrm{~g}}\)
= 0.556 mol kg^-1

Question 4.
If N₂ gas is bubbled through water at 293 K, how many mililmoles of N₂ gas would dissolve in 1 litre of water ? Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henrys law constant for N₂ at 293 K is 76.48 k bar.
Solution:
The solubiity of gas is related to the mole fraction in aqueous solution.
The mole fraction of the gas in the solution is calculated by applying Henrys law. Thus:
x (nitrogen) = \(\frac{\mathrm{p} \text { (nitrogen) }}{\mathrm{K}_{\mathrm{H}}}=\frac{0.987 \mathrm{bar}}{76.480 \mathrm{bar}}\) = 1.29 × 10^-5
As 1 Litre of water contains 55.5 mol of it, therefore if n represents number of moles of N₂ in solution.
x(Nitrogen) = \(\frac{\mathrm{n} \mathrm{mol}}{\mathrm{n} \mathrm{mol}+55.5 \mathrm{~mol}}=\frac{\mathrm{n}}{55.5}\) = 1.29 × 10^-5
(n in denominator is neglected as it is << 55.5)
Thus n = 1.29 × 10^-5 55.5 mol 7.16 × 10^-4 mol
= \(\frac{7.16 \times 10^{-4} \mathrm{~mol} \times 1000 \mathrm{mmol}}{1 \mathrm{~mol}}\) = 0.716 m mol

Question 5.
Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K and (ii) mole fractions of each component in vapour phase.
Solution:
i) Molar mass of CH₂Cl₂ = 12 × 1 + 1 × 2 + 35.5 × 2 = 85 g mol^-1
Molar mass of CHCl₃ = 12 × 1 + 1 × 1 +35.5 × 3 = 119.5 g mol^-1
Moles of CH₂Cl₂ = \(\frac{40 \mathrm{~g}}{85 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.47 mol
Moles of CHCl₃ = \(\frac{25.5 \mathrm{~g}}{119.5 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.2 13 mol
Total number of moles = 0.47 + 0.2 13 = 0.683 mol
X_CH2Cl2 = \(\frac{0.47 \mathrm{~mol}}{0.683 \mathrm{~mol}}\) = 0.688
X_CHCl3 = 1.00 – 0.688 = 0.3 12
Using equation :
p_total = \(\mathrm{p}_{1}^{0}+\left(\mathrm{p}_{2}^{0}-\mathrm{p}_{1}^{0}\right)\) x₂ = 200 + (415 – 200) × 0.688
= 200 + 147.9 = 347.9mm Hg

ii) Using the relation, y₁ = p₁/p_total we can calculate the mole fraction of the components in gas phase (y₁)
P_CH2Cl2 = 0.688 × 415 mm Hg = 285.5 mm Hg
P_CHCl3 = 0.3 12 × 200 mm Hg = 62.4 mm Hg
y_CH2Cl2 = 285.5 mm Hg/347.9 mm Hg = 0.82
y_CHCl3 = 62.4 mm Hg/347.9 mm Hg = 0.18

Note : Since CH₂Cl₂ is a more volatile component than CHCl₃, [\(\mathrm{p}_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}^{0}\) = 415 mm Hg and
\(\mathrm{p}_{\mathrm{CHCl}_{3}}^{0}\) = 200 mm Hg] and the vapour phase is also richer in CH₂Cl₂ [y_CH2Cl2 = 0.82 and y_CHCl3 = 0.18], it may thus be concluded that at equilibrium, vapour phase will be always rich in the component which is more volatile.

Question 6.
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0 g of benzene (molar mass 78 g mol^-1), vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance ? [A.P. Mar. 18; 16]
Solution:
The various quantities known to us are as follows.
\(\mathrm{p}_{1}^{0}\) = 0.850 bar
P = 0.845bar
M₁ = 78 g mol^-1
w₂ = 0.5 g
w₁ = 39 g
Substituting these values in equation \(\frac{\mathrm{P}^{0}-\mathrm{P}}{\mathrm{P}_{1}^{0}}=\frac{\mathrm{w}_{2} \times \mathrm{M}_{1}}{\mathrm{M}_{2} \times \mathrm{w}_{1}}\) we get
\(\frac{0.850 \mathrm{bar}-0.845 \mathrm{bar}}{0.850 \mathrm{bar}}=\frac{0.5 \mathrm{~g} \times 78 \mathrm{~g} \mathrm{~mol}^{-1}}{\mathrm{M}_{2} \times 39 \mathrm{~g}}\)
Therefore, M₂ = 170 g mol^-1.

Question 7.
18g of glucose, C₆H₁₂O₆, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar ? K_b for water is 0.52 K kg mol^-1.
Solution:
Moles of glucose = 18g/180 g mol^-1 = 0.1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0.1 mol kg^-1
For water, change in boiling point
∆T_b = K_b × m = 0.52 K kg mol^-1 × 0.1 mol kg^-1 = 0.052 K .
Since water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K.

Question 8.
The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_b for benzene is 2.53 K kg mol^-1.
Solution:
The elevation of (∆T_b) in.the boiling point = 354.11 K – 353.23 K. = 0.88 K
Substituting these values in expression we get, M₂ = \(\frac{1000 \times \mathrm{w}_{2} \times \mathrm{K}_{\mathrm{b}}}{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{w}_{1}}\)
M₂ = \(\frac{2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1.8 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{0.88 \mathrm{~K} \times 90 \mathrm{~g}}\)
Therefore, molar mass of the solute, M₂ = 58 g mol^-1.

Question 9.
45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.
Solution:
Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol = \(\frac{\text { moles of ethylene glycol }}{\text { mass of water in kilogram }}\)
Moles of ethylene glycol = \(\frac{45 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.73 mol
Mass of water in kg = \(\frac{600 \mathrm{~g}}{1000 \mathrm{~g} \mathrm{~kg}^{-1}}\) = 0.6 kg
Hence molality of ethylene glycol = \(\frac{0.73 \mathrm{~mol}}{0.60 \mathrm{~kg}}\) = 1.2 mol kg^-1
Therefore freezing point depression,
∆T_f = 1.86 K kg mol^-1 × 1.2 mol kg^-1 = 2.2 K
Freezing point of the aqueous solution 273.15 K – 2.2 K = 270.95 K

Question 10.
1.00 g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^-1. Find the molar mass of the solute.
Solution:
Substituting the values of various terms involved in equation
M₂ = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{2} \times 1000}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{w}_{1}}\) we get
M₂ = \(\frac{5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1.00 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{0.40 \times 50 \mathrm{~g}}\) = 256 g mol^-1
Thus, molar mass of the solute = 256 g mol^-1

Question 11.
200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotiç pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar mass of the protein.
Solution:
The various quantities known to us are as follows : π = 2.57 × 10^-3 bar.
V = 200 cm³ = 0.200 litre
T = 300K
R = 0.083 L bar mol^-1 K^-1
Substituting these values in equation,
we get M₂ = \(\frac{w_{2} R T}{\pi V}\)
M₂ = \(\frac{1.26 \mathrm{~g} \times 0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{2.57 \times 10^{-3} \mathrm{bar} \times 0.200 \mathrm{~L}}\)
= 61.022 g mol^-1

Question 12.
2 g of benzoic acid (C₆H₅COOH) dissolved in 25g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol^-1. What is the percentage association of acid if it forms dimer in solution ?
Solution:
The given quantities are :
w₂ = 2g; K_f = 4.9 K kg mol^-1; w₁ = 25 g
∆T_f = 1.62 K
Substituting these values, in equation, we get:
M₂ = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{2} \times 1000}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{w}_{1}}\)
M₂ = \(\frac{4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 2 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{25 \mathrm{~g} \times 1.62 \mathrm{~K}}\) = 241.98 g mol^-1
Thus, experimental molar mass of benzoic acid in benzene is = 241.98 g mol^-1
Now consider the following equilibrium for the acid :
2 C₆H₅COOH ⇌ (C₆H₅COOH)₂
If x represents the degree of association of the solute then we would have (1 – x) mol of benzoic acid left in unassociated form and correspondingly \(\frac{x}{2}\) as associated moles of benzoic acid at equilibrium.
Therefore, total number of particles at equilibrium is :
1 – x + \(\frac{x}{2}\) = 1 – \(\frac{x}{2}\).
Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.
But i = \(\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}=\frac{122 \mathrm{~g} \mathrm{~mol}^{-1}}{241.98 \mathrm{~g} \mathrm{~mol}^{-1}}\)
or = \(\frac{x}{2}\) = 1 – \(\frac{122}{241.98}\) = 1 – 0.504 = 0.496
or x = 2 × 0.496 = 0.992
Therefore, degree of association of benzoic acid in benzene is 99.2%.

Question 13.
0.6 mL of acetic acid (CH₃COOH), having density 1.06 g mL^-1, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.
Answer:
Number of moles of acetic acid = \(\frac{0.6 \mathrm{~mL} \times 1.06 \mathrm{~g} \mathrm{~mL}^{-1}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.0106 mol = n
Molality = \(\frac{0.0106 \times \mathrm{mol}}{1000 \mathrm{~mL} \times 1 \mathrm{~g} \mathrm{~mL}^{-1}}\) = 0.0106 mol kg^-1
Using equation ∆T_f = \(\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{w}_{1}}\)
∆T_f = 1.86 K kg mol^-1 × 0.0106 mol kg^-1 = 0.0197 K
van’t Hoff factor (i) = \(\frac{\text { Observed freezing point }}{\text { Calculated freezing point }}=\frac{0.0205 \mathrm{~K}}{0.0197 \mathrm{~K}}\) = 1.041
Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen ions per molecule of acetic acid. If x is the degree of dissociation of acetic acid, then we would have n (1 — X) moles of undissociated acetic acid, mc moles of CH₃COO^– and nx moles of H⁺ ions.

Intext Questions

Question 1.
Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachlodie (CCl₄) if 22g of benzene is dissolved in 122g of carbon tetrachioride.
Then, calculate the mass percentage from the formula
Mass % = × 100
Solution:
Mass of benzene = 22g; Mass of CCl₄ = 122g
Mass of solution = 22 + 122 = 144g
Mass % of benzene = \(\frac{22}{144}\) × 100 = 15.28%
Mass of CCl₄ = 100 – 15.28 = 84.72%
Note: Mass percent of CCl₄ can also be calculated by using the formula as:
Mass % of CCCl₄ = \(\frac{122}{144}\) × 100 = 84.72%

Question 2.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
• Then calculate the mole fraction by using the formula
Mole fraction of a component = \(\frac{\text { Number of moles of the component }}{\text { Total number of moles of all components }}\)
x_A = \(\frac{\mathbf{n}_{\mathbf{A}}}{\mathbf{n}_{\mathbf{A}}+\mathbf{n}_{\mathbf{B}}}\)
Solution:
For 100 g of the solution
Mass of benzene = 30 g
Mass of carbon tetrachioride = 100 – 30 = 70g
Molar mass of benzene (C₆H₆) (12 × 6) + (6 × 1) = 72 + 6 = 78 g mol^-1
Moles of benzene, n_C6H6 = \(\frac{\text { Mass }}{\text { Molar mass }}=\frac{30}{78}\) = 0385 mol
Molar mass of carbon tetrachioride (CCl₄) = 12 + (35.5 × 4)
= 12 + 142.0 = 154 g mol^-1
Moles of CCl₄,n_CCl4 = \(\frac{70 \mathrm{~g}}{\left(154 \mathrm{~mol}^{-1}\right)}\) = (154 mol^-1)
Mole fraction of benzene, X_C6H6 = \(\frac{\mathrm{n}_{\mathrm{C}_{6} \mathrm{H}_{6}}}{\mathrm{n}_{\mathrm{C}_{6} \mathrm{H}_{6}}+\mathrm{n}_{\mathrm{CC}_{4}}}=\frac{0.385 \mathrm{~mol}}{(0.385+0.454) \mathrm{mol}}\) = 0.459

Question 3.
Calculate the molarity of each of the following solution:
a) 30g of CO(NH₃)₂. 6H₂O in 4.3 L of solution.
b) 30 mL of 0.5 MH₂SO₄ diluted to 500 mL.

So, first find molar mass by adding atomic masses of different elements, then find moles of solute and then molarity.
b)Use molarity equation for dilution.
M₁V₁
(Before dilution)
M₂V₂
(After dilution)
Solution:
a) Molar mass of CO(NO₃)₂. 6H₂O = (58.7) + 2(14 + 48) + (6 × 18) g mol^-1
= 58.7 + 124 + 108 = 290.7 = 291g mol^-1
Mole of CO(NO₃)₂ 6 H₂O = \(\frac{30 \mathrm{~g}}{291 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.103
Volume of solution = 4.3 L
Molarity (M) = \(\frac{0.103 \mathrm{~mol}}{4.3 \mathrm{~L}}\) = 0.024 mol L^-1 = 0.024 M

b) Volume of undiluted H₂SO₄ solution (V₁) = 30 mL
Molarity of undiluted H₂SO₄ solution (M₁) = 0.5 M
Volume of diluted H₂SO₄ solution (V₂) = 500 mL
We know that M₁V₁ = M₂V₂
∴ M₂ = \(\frac{\mathrm{M}_{1} \mathrm{~V}_{1}}{\mathrm{~V}_{2}}=\frac{(0.5 \mathrm{M})(30 \mathrm{~mL})}{500 \mathrm{~mL}}\) = 0.03 M

Question 4.
Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molar aqueous solution.

So, find the molar mass of solute by adding atomic masses of different element present in it and mass by using the formula,
Molality =
Mass of solvent in kg
Solution:
Molality of the solution = 0.25 m = 0.25 mol kg^-1
Molar mass of urea (NH₂CONH₂) = (14 × 2) + (1 × 4) + 12 + 16
= 60 g mol^-1
Mass of solvent (water) = 2.5 kg
Molality =
(0.25 mol kg^-1) = \(\frac{\text { Mass of urea }}{\left(60 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(2.5 \mathrm{~kg})}\)
Mass of urea = (0.25 mol kg^-1) × (60 g mol^-1) × (2.5 kg) = 37.5 g

Question 5.
Calculate a) molality b) molarity and c) mole fraction of KI if the density of 20% (mass / mass) aqueous Kl is 1.202 g mL^-1.
• As density and % by mass is given, so find the mass of solute and solvent (as x % solution contains x g solute in (100 – x) g solvent).
Find volume of the solution, by using,
Volume = \(\frac{\text { Mass }}{\text { Density }}\)
Recall the formulae of molality, molarity and mole fraction, to calculate them.
Molality =
Solution:
a) Molality
Weight of KI in 100 g of water = 20 g
Weight of water in the solution = 100 – 20 = 80 g = 0.08 kg
Molar mass of KI = 39 + 127 = 166 g mol^-1
Molarity of the solution (m) = \(\frac{\text { Number of moles of KI }}{\text { Mass of water in } \mathrm{kg}}\)
= \(\frac{(20 \mathrm{~g}) /\left(166 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{(0.08 \mathrm{~kg})}\) = 1.506 mol kg^-1 = 1.506 m

b) Molarity
Weight of the solution = 100 g
Density of the solution = 1.202g m^-1

c) Mole fraction of KI

Question 6.
H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195m, calculate Henry’s law constant.
Solution:
a) Calculation of mole fraction of H₂S 0.195 m means that 0.195 mole of H₂S are dissolves in 1000 g of water.
Number of moles of water in 1000g, (n_H2O) = \(\frac{(1000 \mathrm{~g})}{\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 55.55 mol
Mole fraction H₂S (x_H2S) = \(\frac{\mathrm{n}_{\mathrm{H}_{2} \mathrm{~s}}}{\mathrm{n}_{\mathrm{H}_{2} \mathrm{~s}}+\mathrm{n}_{\mathrm{H}_{2} \mathrm{O}}}\)
= \(\frac{(0.195 \mathrm{~mol})}{(0.195+55.55) \mathrm{mol}}=\frac{(0.195 \mathrm{~mol})}{(55.745 \mathrm{~mol})}\) = 0.0035

b) Calculation of Henry’s law constant:
According to Henry’s law

Question 7.
Henrys law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at’298 K.
Solution:
Step I : Calculation of number of moles of CO₂
According to Henrys law,

n_CO2 = x_CO2 × (27.78 mol) (1.52 10^-3) × (27.78 mol) = 0.0422 mol
Note: n_CO2 is considered negligible due to its little solubility in water.
Step II : Calculation of mass of dissolved CO₂ in water
Mass of CO₂ = No. of moles CO₂ × Molar mass of CO₂
= (0.0422 mol) × (44 g mol^-1) = 1.857 g

Question 8.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
. Apply Raoults law p_T = \(\mathbf{p}_{\mathrm{A}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{B}}=\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0}\left(1-\mathbf{x}_{\mathrm{A}}\right)\) to calculate mole fraction of A(x_A) and B(x_B).
In vapour phase, partial pressure are used instead of number of moles.
Solution:
Step I: ComposItion In liquid phase
Vapour pressure of pure liquid A (\(p_{A}^{0}\)) = 450 mm
Vapour pressure of pure liquid B (\(p_{B}^{0}\)) = 700 mm
Total vapour pressure of the solution (p) = 600 mm
According to Raoult’s law,
p = \(\mathbf{p}_{\mathrm{A}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{B}}=\mathbf{p}_{\mathrm{B}}^{0} \mathbf{x}_{\mathrm{A}}+\mathbf{p}_{\mathrm{B}}^{0}\left(1-\mathbf{x}_{\mathrm{A}}\right)\)
(600 mm) = 450 mm × x_A + 700 mm (1 – x_A)
=700 mm + x_A + (450 – 700)mm
= 700 – x_A (250 mm)
x_A = \(\frac{(600-700)}{-(250 \mathrm{~mm})}\) = 0.40
Mole fraction of A(x_A) = 0.40
Molefraction of B(x_B) = 1 – 0.40 = 0.60

Step II: Composition in vapour phase
P_A = \(\mathrm{p}_{\mathrm{A}}^{0} \mathrm{x}_{\mathrm{A}}\) = (450 mm) × 0.40 = 180 mm
P_B = \(\mathrm{p}_{\mathrm{B}}^{0} \mathrm{x}_{\mathrm{B}}\) = (700 mm) × 0.60 = 420 mm
Mole fraction of A in vapour phase = \(\frac{\mathrm{p}_{\mathrm{A}}}{\mathrm{p}_{\mathrm{A}}+\mathrm{p}_{\mathrm{B}}}\)
= \(\frac{(180) \mathrm{mm}}{(180+420) \mathrm{mm}}\) = 0.30
Mole fraction of B in vapour phase = \(\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}+\mathrm{p}_{\mathrm{B}}}=\frac{(420) \mathrm{mm}}{(180+420) \mathrm{mm}}\) = 0.70

Question 9.
Vapour pressure of pure water at 298 K Is 23.8 mm Hg. 50g urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and Its relative lowering.
Consider Raoult’s law and formula for relative lowering in vapour pressure,
\(\frac{\mathbf{p}_{\mathrm{A}}^{0}-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathrm{A}}^{0}}=\frac{\mathbf{n}_{\mathrm{B}}}{\mathbf{n}_{\mathrm{A}}}=\frac{\mathbf{W}_{\mathrm{B}}}{\mathbf{M}_{\mathrm{B}}} \times \frac{\mathbf{M}_{\mathrm{A}}}{\mathbf{W}_{\mathrm{A}}}\)
Where, \(\frac{\mathbf{p}_{\mathrm{A}}^{0}-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathrm{A}}^{0}}\) is called relative lowering in vapour pressure.
Solution:
Step I : Calculation of vapour pressure of water for this solution.
According to Raoult’s law,
\(\frac{\mathrm{p}_{\mathrm{A}}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^{0}}=\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{n}_{\mathrm{A}}}=\frac{\mathrm{W}_{\mathrm{B}} / \mathrm{M}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} / \mathrm{M}_{\mathrm{A}}}=\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \times \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{A}}}\) ………………. (1)
(Pure water) \(\mathrm{p}_{\mathrm{A}}^{0}\) = 23.8 mm;
W_B (urea) = 50 g; W_A (water) = 850 g
M_B (urea) = 60 g mol^-1; M_A (water) = 180 g mol^-1
Placing the values in eq. (i)
\(\frac{\mathrm{p}_{\mathrm{A}}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^{0}}=\frac{(50 \mathrm{~g}) \times\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{\left(60 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(850 \mathrm{~g})}=0.01762\)
\(\frac{23.8-\mathrm{p}_{\mathrm{s}}}{238}\) = 0.01762; 23.8 – p_s = 0.4194
p_s = 23.3806 ≈ 23.38 mm Hg

Step II : Calculation of relative lowering of vapour pressure
Relative lowering in vapour pressure = \(\frac{\mathrm{p}_{\mathrm{A}}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^{0}}=\frac{(23.8-23.38) \mathrm{mm}}{(23.8 \mathrm{~mm})}\) = 0.0176

Question 10.
Boiling point of water 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
[K_b for water is 0.52 K kg mol^-1]
i) Since boiling point is changing, apply the formula for elevation In boiling point,
∆T_b = K_bm

iii) Find ∆T_b as ∆T_b = T_b = T_b – \(T_{b}^{0}\)
T_b = Boiling point of solution
T_b = Boiling point of pure solvent
Solution:
Mass of water (W_A) = 500 g = 0.5 kg
Elevation in boiling point (∆T_b) = 100°C – 99.63°C = 037°C = 0.37 K
Molal elevation constant (K_b) = 0.52 K kg mol^-1
Molar mass sucrose C₁₂H₂₂O₁₁
(M_B) = (12 × 12) + (22 × 1) + (16 × 11)
= 342 g mol^-1
W_B = \(\frac{\mathrm{M}_{\mathrm{B}} \times \Delta \mathrm{T}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{b}}}\)
= \(\frac{\left(342 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.37 \mathrm{~K}) \times(0.5 \mathrm{~kg})}{\left(0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}\) = 121.7 g

Question 11.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol^-1.
. Since, lowering of melting point is given apply the formula for lowering of melting point, i.e.,
∆T_f = K_f.m
∆T_f = \(\frac{\mathbf{K}_{f} \cdot \mathbf{W}_{\mathbf{B}}}{\mathbf{M}_{\mathbf{B}} \times \mathbf{W}_{\mathbf{A}}}\) or W_B = \(\frac{\Delta \mathbf{T}_{\mathrm{f}^{*}} \mathbf{M}_{\mathrm{B}} \cdot \mathbf{W}_{\mathrm{A}}}{\mathbf{K}_{\mathrm{f}}}\)
Solution:
Mass of ascorbic acid (W_A) = 75 g = 0.075 kg
Depression in melting point (∆T_f) = 1.5° C = 1.5 K
Molar mass of ascorbic acid (M_B) = (12 × 6) + (8 × 1) + (16 × 6) = 176 g mol^-1
Molal depression constant (K_f) = 3.9 K kg mol^-1
W_B = \(\frac{\left(176 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(1.5 \mathrm{~K}) \times(0.075 \mathrm{~kg})}{\left(3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}\) = 5.08 g

Question 12.
Calculate the osmotic pressure in Pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 1,85,000 in 450 mL of water at 37°C.
Use the formula for osmotic pressure (π) = CRT and C = \(\frac{\mathbf{n}}{\mathbf{V}}\) and n = \(\frac{\mathbf{W}_{\mathbf{B}}}{\mathbf{M}_{\mathbf{B}}}\)
Solution:
Mass of polymer (W_B) = 1.0 g
Molar mass of polymer (M_B) = 185000 g mol^-1
Volume of solution (V) = 450 mL = 0.450 L
Temperature (T) = 37 + 273 = 310 K
Solution constant (R) = 8.314 × 10³ Pa L K^-1 mol^-1
Osmotic pressure (π) = CRT
= \(\frac{\mathrm{W}_{\mathrm{B}} \times \mathrm{R} \times \mathrm{T}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{V}}\)
π = \(\frac{(1.0 \mathrm{~g}) \times\left(8.314 \times 10^{3} \mathrm{~Pa} \mathrm{~L} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(310 \mathrm{~K})}{\left(185000 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.450 \mathrm{~L})}\)
= 30.96 Pa

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 1st Lesson Solid State Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 1st Lesson Solid State

Very Short Answer Questions

Question 1.
Define the term amorphous.
Answer:
An amorphous (no form) solid is a compound which does not have an orderly arrangement of particles. In amorphous solids the constituent particles, atoms, molecules has short-range order only.
E.g.: Glass, rubber, plastics, etc.,

Question 2.
What makes a glass different from quartz?
Answer:

• Glass is an amorphous solid in which the constituent particles have only short-range order.
• Quartz is a crystalline form of silica in which constituent particles have long-range order.

Question 3.
Classify the following solids as ionic, metallic, molecular, covalent network or amorphous.
i) Si
ii) I₂
iii) P₄
iv) Rb
v) SiC
vi) LiBr
vii) Ammonium .Phosphate (NH₄)₃ PO₄
viii) Plastic
ix) graphite
x) Tetra phosphorous decoxide
xi) brass
Answer:
i) Si – Covalent network solid
ii) I₂ – Molecular solid with covalent bonds
iii) P₄ – Molecular solid with covalent bonds ,
iv) Rb – Metallic solid
v) SiC – Giant molecular Network Solid with covalent bonds
vi) LiBr – Ionic solids
vii) Ammonium Phsophate (NH₄)₃ PO₄ – Ionic Solids
viii) Plastic – Amorphous solid
ix) Graphite – Hexagonal network solid with covalent bonds (giant molecule)
x) Tetra phosphorous decoxide (P₄O₆) – Molecular solid with covalent bonds.
xi) Brass – Metallic solid

Question 4.
What is meant by the term coordination number ?
Answer:
The number of nearest neighbouring particles of a particle is defined as the co-ordination number.
(Or)
The number of nearest oppositely charged ions surrounding a particular ion is also called as co-ordination number.
E.g.: Co-ordination no. of Na⁺ in NaCl lattice is ‘6’.

Question 5.
What is the co-ordination number of atoms in a cubic close – pack structure ?
Answer:
The co-ordination number of atoms in a cubic close pack structure is ’12’.

Question 6.
What is the co-ordination number of atoms in a body – centered cubic structure ?
Answer:
The co-ordination number of atoms in a body – centered cubic structure is ‘8’.

Question 7.
Stability of a crystal is reflected in the magnitude of its melting point. Comment.
Answer:
Stability of a crystal is reflected in the magnitude of its melting point.
Explanation :

• The stability of a crystal mainly depends upon the magnitude of forces of attraction between the constituent particles.
• As the attractive forces between the constituent particles increases stability of the crystal aslo increases.
• As the stability of crystal increases melting point of solid will be higher.

Question 8.
How are the intermolecular forces amonj the molecules affect the melting point ?
Answer:

• As the intermolecular forces between constituent particles of solid increases, stability of the compound increases.
• As the stability of crystal increases belting point of solid also increases (high).

Question 9.
How do you distinguish between hexagonal close – packing and cubic close – packing structures ?
Answer:
Hexagonal close packing : The spheres of the 3^rd layer are exactly aligned with those of first layer. This pattern of spheres is repeated in alternate layers. Tetrahedral voids of the 2^nd layer may be covered by the spheres of 3^rd layer. This structure is called hexagonal close packed (hep) structure.
Cubic close packing : The spheres of 3^rd layer cover the octahedral voids of 2^nd layer. But the spheres of 4th layers are aligned with those of first layer. This structure is called cubic close packing.

Question 10.
How do you distinguish between crystal lattice and unit cell ?
Answer:
Crystal lattice : A regular arrangement of the constituent particles of a crystal in the three dimensional space is called crystal lattice.
Unit cell: The simple unit of crystal lattice which when repeated again and again gives the entire crystal of a given substance is called unit cell.

Question 11.
How many lattice points are there in one unit cell of face centered cubic lattice ?
Answer:
In face centered cubic unit cell.
The number of comer atoms per unit cell.
= 8 comer atoms × \(\frac{1}{8}\) atom per unit cell = 8 × \(\frac{1}{8}\) =1 atom
Number of face centered atoms per unit cell
= 6 face centered atoms × \(\frac{1}{2}\) atom per unit cell
= 6 × \(\frac{1}{8}\) = 3 atoms
∴ Total no. of lattice points = 1 + 3 = 4.

Question 12.
How many lattice points are there in one unit cell of face – centered tetragonal lattice ?
Answer:
In face centered tetragonal unit cell
Number of face centered atoms per unit cell
= 6 face centered atoms × \(\frac{1}{2}\) atom per unit cell
= 6 × \(\frac{1}{2}\) = 3 atoms
∴ Total no. of lattice points = 1 + 3 = 4.

Question 13.
How many lattice points are there in one unit cell of body centered cubic lattice ?
Answer:
In body – centered cubic unit cell
The number of comer atoms per unit cell
= 8 comers × \(\frac{1}{8}\) per comer atom
= 8 × \(\frac{1}{8}\) = 1 atom
Number of atoms at body center = 1 × 1 = 1 atom
∴ Total no. of lattice points = 1 + 1 = 2.

Question 14.
What is a semi conductor ?
Answer:
The solids which are having moderate conductivity between insulators and conductors are called semi conductors.
These have the conductivity range from 10^-6 to 10⁴ Ohm^-1 m^-1.
By doping process the conductivity of semi conductors increases.
E.g. : Si, Ge crystal.

Question 15.
What is Schottky defect ?
Answer:
Schottky defect:

1. It is a point defect in which an atom or ion is missing from its normal site in the lattice”.
2.  In order to maintain electrical neutrality, the number of missing cations and anions are equal.
3. This sort of defect occurs mainly in highly ionic compounds, where cationic and anionic sizes are similar.
   In such compounds the co-ordination number in high.
   Ex.: NaCl, CsCl etc.
4. Illustration:
   
5. This defect decreases the density of the substance.

Question 16.
What is Frenkel defect ?
Answer:
Frenkel defect:

1. “It is a point defect in which an atom or ion is shifted from its normal lattice position”. The ion or the atom now occupies an interstitial position in the lattice.
2. This type of a defect is favoured by a large difference in sizes between the cation and anion. In these compounds co-ordination number is low.
   E.g.: Ag – halides, ZnS etc.
3. Illustration:
   
4. Frenkel defect do not change the density of the solids significantly.

Question 17.
What is interstitial defect ?
Answer:

• Some of the constituent particles of solid compound occupy an interstitial site, the crystal is said to have interstitial defect.
• This defect shown by ionic solid, non-ionic solids which are maintaining electrical neutrality.
  

Question 18.
What are f – centers ?
Answer:

• f – centers are the anionic sites occupied by unpaired electrons.
• These impart colour to cyrstals. This colour is due to the excitation of electrons when they absorb energy from the visible light.
• f – centres are formed by heating alkyl halide with excess of alkali metal.
  E.g.: NaCl crystals heated in presence of Na – vapour yellow colour is produced due to f – centres.

Question 19.
Explain Ferromagnetism with suitable example.
Answer:
Ferromagnetic Substances : Some substances containing more number of unpaired electrons are very strongly attracted by the external magrietic field. In Ferromagnetic substances the Magnetic moments in individual atoms are all alligned in the same direction. Such substances are called Ferromagnetic Substances. In ferromagnetic substances the field strength B > > > H.
E.g. : Fe, Co and Ni.

Question 20.
Explain paramagnetism with suitable example.
Answer:
Paramagnetic Substances : If the magnetic lines of forces are drawn into the substance the field (B) in the substances is greater than the applied field (H) i.e., B > H. Such a substance is called paramagnetic substance. Paramagnetic substance moves from a weaker part of the field to a stronger part of the field. Paramagnetic substances are weakly attracted in a external Magnetic field. They exhibit paramagnetism due to the presence of unpaired electrons.
E.g. : Cr⁺³, Sc⁺², K₃ [Fe (CN)₆].

Question 21.
Explain Ferrimagnetisms with suitable example.
Answer:
Ferrimagnetism is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti parallel directions in unequal numbers.

• These are weakly attracted by magnetic field as compared to ferromagnetic substances.
• These lose ferrimagnetism on heating and becomes paramagnetic.

Question 22.
Explain Antiferromagnetism with suitable example.
Answer:
Substances like Mno showing anti-ferromagnetism having domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each others magnetic moment.

Question 23.
Why X – rays are needed to probe the crystal structure ?
Answer:
According to the principles of optics, the wavelength of light used to observe an object must be no greater than the twice the length of the object it self. It is impossible to see atom s using even the finest optical microscope. To see the atoms we must use light with a wavelength of approximately 10^-10 m. X – rays are present with in this region of electromagnetic spectrum. So X – rays are used to probe crystal structure.

Short Answer Questions

Question 1.
Explain similarities and differences between metallic and ionic crystals.
Ans:
Similarities between ionic and metallic crystals :

• Both ionic and metallic crystals have electrostatic force of attraction.
• The bond present in ionic crystals and metallic crystals is non-directional.

Differences between ionic and metallic crystals :

Ionic Crystals

1. In ionic crystals electrostatic force is in between oppositely charged ions.
2. These are good conductors of electricity.
3. Ionic bond in ionic crystals is strong.

Metallic Crystals

1. In metallic crystals electrostatic force is in between the valency electrons.
2. These are goods conductors of electricity  in solid state.
3.  Metallic bond in metallic crystals is weak (or) strong.

Question 2.
Explain why ionic solids are hard and brittle.
Answer:
In ionic solids the formation of solid compound is due to the arrangements of cations and anions bound by strong coloumbic force i.e., electro static force. So ionic solids are hard and brittle in nature. These have high melting and boiling points.

Question 3.
Calculate the efficiency of packing in case of a metal of simple cubic crystal.
Answer:
Packing efficiency in case of metal of simple cubic crystal:

The edge length of the cube
a = 2r. (r = radius of particle)
Volume of the cubic unit cell = a³ = (2r)³
= 8r³
∵ A simple cubic unit cell contains only one atom
The volume of space occupied = \(\frac{4}{3}\) πr³
∴ Packing efficiency
= \(\frac{\text { Volume of one atom }}{\text { Volume of cubic unit cell }}\) × 100
= \(\frac{4 / 3 \pi r^{3}}{8 r^{3}}\) × 100 = \(\frac{\pi}{6}\) × 100 = 52.36%.

Question 4.
Calculate the efficiency of packing in case of a metal of body centered cubic crystal.
Answer:
Packing efficiency in case of a metal of body centred cubic crystal:

in B.C.C. Crystal
\(\sqrt{3}\)a = 4r
a = \(\frac{4 \mathrm{r}}{\sqrt{3}}\)
In this structure total no. of atoms is ‘2’ and their volume = 2 × (\(\frac{4}{3}\)) πr³
Volume of the cube = a³ = (\(\frac{4}{\sqrt{3}}\)r)³

Question 5.
Calculate the efficiency of the packing incase of face – centered cubic crystal.
Answer:
Packing efficiency of face centered cubic crystal:
In fcc lattice assume that atoms are touching each other. Here unit cell edge length
a = 2\(\sqrt{2}\) r
Each unit cell as effectively four spheres.
Total volume of four spheres = 4 × (\(\frac{4}{3}\)) πr³
Volume of cube = a³ = (2\(\sqrt{2}\)r)³

Question 6.
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body – centre. What is the formula of the compound ? What are the coordination numbers of P and Q?
Answer:
The contribution of atoms Q present at 8 corners of the cube = \(\frac{1}{8}\) × 8 = 1
The contribution of atoms P present at the body centre = 1
Therefore ratio of P and Q = 1 : 1
So, the formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

Question 7.
If the radius of the octahedral void is ‘r and radius of the atoms in close packing is ‘R’, derive relation between r and R.
Answer:
Derivation of relation between r and R:
Octahedral void is shown in the following diagram as a shaded circle.

∴ ∆ ABC is a right angled triangle.
∴ We apply pythagoras theorem.
AC² = AB² + BC²
(2R)² = (R + r)² + (R + r)² = 2 (R + r)²
4R² = 2 (R + r)²
2R² = (R + r)²
(\(\sqrt{2}\) R)P = (R + r)²
\(\sqrt{2}\) R = R + r
r = \(\sqrt{2}\) R – R
r = (\(\sqrt{2}\) – 1)R
r = (1.414- 1) R
r = 0.414 R.

Question 8.
Describe the two main types of semiconductors and contrast their conduction mechanism. [A.P. Mar. 19]
Answer:
The solids which are having moderate conductivity between insulators and conductors are called semi conductors.
These have the conductivity range from 10^-6 to 10⁴ Ohm^-1m^-1.
By doping process the conductivity of semi conductors increases. E.g.: Si, Ge, crystal.
Semi conductors are Of two types. They are :
1. Intrinsic semi-conductors : In case of semi-conductors, the gap between the valence band and conduction band is small. Therefore, some electrons may jump to conduction band and show some conductivity. Electrical conductivity of semi-conductors increases with rise in “temperature”, since more electrons can jump to the conduction band. Substances like silicon and germanium show this type of behaviour and are called intrinsic semi-conductors.

2. Extrinsic semi – conductors : Their conductivity is due to the presence of impurities.
They are formed by “doping”. ,
Doping: Conductivity of semi-conductors is too low to be of pratical use. Their conductivity is increased by adding an appropriate amount of suitable impurity. This process is called “doping”.
Doping can be done with an impurity which is electron rich or electron deficient.

Extrinsic semi-conductors are of two types.
a) n-type semi-conductors : It is obtained by adding trace amount of V group element (P, As, Sb) to pure Si or Ge by doping.
When P, As, Sb (Or) Bi is added to Si or Ge some of the Si or Ge in the crystal are replaced by P or As atoms and four out of five electrons of P or As atom will be used for bonding with Si or Ge atoms while the fifth electron serve to conduct electricity.

b) p-type semi-conductors : It is obtained by doping with impurity atoms containing less electrons i.e., III group elements (B, Al, Ga or In).
When B or Al is added to pure Si or Ge some of the Si or Ge in the crystal are replaced by B or AZ atoms and four out of three electrons of B or Al atom will be used for bonding with “Si” or Ge atoms while the fourth valence electron is missing is called electron hole (or) electron vacancy. This vacancy on an atom in the structure migrates from one atom to another. Hence it facilitates the electrical conductivity.

Question 9.
Classify each of the following as either a p – type or a n – type semiconductor.
1. Ge doped with In
2. Si doped with B.
Answer:
Both (1) and (2) come under “p – type semiconductors”.
Reason: In both the cases dopants (i.e.,) Indium in case – (1) and Boron in case – (2) belong to III (or) 13^th group. Si (or) Ge doped with III group element is known as p-type semi- conductor.
Explanation : Doping the silicon or germanium element with III or 13^th group element like “B”, Al, Ga or “In” results in the substitution of some silicon atoms in its structure by the dopant. The dopant has only three valency electrons. The fourth valency electron is required. It is left as a vacant place on the atom. It is known as an ‘electron vacancy’ (or) a ‘hole’. The electron vacancy on ah atom in the structure migrates from one atom to another. Hence it facilitates the electrical conductivity. Si (or) Ge, doped with elements that create a hole in the structure, is known as p- type semi-conductor.

Question 10.
Analysis shows that nickel oxide has the formula Ni^0.980, 1.00, what fractions of nickel exist as Ni²⁺ and Ni³⁺ ions ?
Answer:
In pure nickel oxide (NiO) the ratio of Ni and O atoms = 1 : 1
Let x be the no. of Ni (II) atoms replaced by Ni (III) atoms in the oxide
∴ Number of Ni (II) atoms present = 0.98 – x
Total charge on Ni atoms = charge on oxygen atom (∵ the oxide is neutral)
2 (0.98 – x) + 3x = 2
1.96 – 2x + 3x = 2
x = 2 – 1.96 = 0.04
No. of Ni (iii) atoms
% of Ni (III) atoms in Nickel oxide = × 100
= \(\frac{0.04}{0.98}\) × 4.01%
% of Ni (II) atoms in nickel oxide = 100 – 4.01 = 95.99%

Question 11.
Gold (atomic radius = 0.144 nm) crystallizes in a face centered unit cell. What is the length of a side of the unit cell ?
Answer:
In a fee unit cell .
The edge length a = 2\(\sqrt{2}\) r
Given r = 0.144 nm
= 2 × 1.414 × 0.144 = 0.407 nm.

Question 12.
In terms of band theory, what is the difference between a conductor and an insulator ?
Answer:
In case of metals (conductors) the atomic orbitals forms molecular orbitals which are close in energy to each other as to form a band called valency band. If this band is partially filled (or) it overlaps with a higher energy unoccupied band called conduction band. Then electrons can flow easily under an applied electric field and the metal shows conductivity.

In case of insulators the gap between filled valence band and the next higher un occupied band is large, electrons cannot jump to it and such a substance has very small conductivity.

Question 13.
In terms of band theory, what is the difference between a conductor and a semiconductor?
Answer:
In case of metals (conductors) the atomic orbitals forms molecular orbitals which are close in energy to each other as to form a band called valency band. If this band is partially filled (or) it overlaps with a higher energy unoccupied band called conduction band. Then electrons can flow easily under an applied electric field and the metal shows conductivity.

In case of semi conductors, the gap between the valence band and conduction band is small. So some of electrons may jump to conduction band and show some conductivity.

Question 14.
If NaCl is doped with 1 × 10^-3 mol percent of SrCl₂, what is the concentration of cation vacancies ?
Answer:
The addition of SrCl₂ to NaCl , each Sr⁺² ion replaces two Na⁺ ions and occupies only one lattice point in place of Na⁺. Due to this one cation vacancy arised.
The number of moles of cation vacancies in 100 moles of NaCl = 1 × 10^-3
The number of moles of cation vacancies in 1 moles NaCl = \(\frac{1 \times 10^{-3}}{100}\) = 10^-5 mole
Total number of cation vacancies = 10^-5 × 6.023 × 10²³ = 6.023 × 10¹⁸.

Question 15.
Derive Bragg’s equation. [T.S. Mar. 19, 16, 15; A.P. Mar. 17, 16, 15] [Mar. 14]
Answer:
Derivation of Bragg’s equation : When X-rays are incident on the crystal or plane, they are diffracted from the lattice points (lattice points may be atoms or ions or molecules). In the crystal the lattice points are arranged in regular pattern. When the waves are diffracted from these points, the waves may be constructive or destructive interference.

The 1^st and 2^nd waves reach the crystal surface. They undergo constructive interference. Then from the figure 1^st and 2^nd rays are parallel waves. So, they travel the same distance till the wave form AD. The second crystal plane
ray travels more than the first by an extra distance (DB + BC) after crossing the grating for it to interfere with the first ray in a constructive manner. Then only they can be in the same phase with one another. If the two waves are to be in phase, the path difference between the two ways must be equal to the wavelength Q.) or integral multiple of it (nλ, where n = 1, 2, 3 )
(i.e.,) nλ = (DB + BC) [where n = order of diffraction]
DB = BC = d sin θ [θ = angle of incident beam,]
(DB + BC) = 2d sin θ [d = distance between the planes]
nλ = 2d sin θ
This relation is known as Bragg’s equation.

Long Answer Questions

Question 1.
How do you determine the atomic mass of an unknown metal If you know Its density and dimension of its unit cell? Explain.
Answer:
Let the, Atomic weight of crystalline substance = M
Avogadros number = N₀
No. of atoms present per unit cell = Z
Density of unit cell or of the substance = ρ
Unit cell length = a
So, volume of the unit cell = a³ (= V)
∴ Then, mass corresponding to each lattice point = M/N₀
∴ Mass of ‘Z lattice points = \(\frac{\mathrm{ZM}}{\mathrm{N}_{0}}\)
∴ Density (p) of unit cell = \(\frac{\text { mass }}{\text { volume }}=\frac{\frac{\mathrm{Z} \cdot \mathrm{M}}{\mathrm{N}_{0}}}{\mathrm{a}^{3}}=\frac{\mathrm{ZM}}{\mathrm{N}_{0} \mathrm{a}^{3}}\)
Atomic weight M = \(\frac{\rho \times \mathrm{a}^{3} \mathrm{~N}_{0}}{\mathrm{Z}}\)

Question 2.
Silver crystallizes in FCC lattice. If edge of the cell is 4.07 × 10^-8 and density is 10.5 g. cm^-3. Calculate the atomic mass of silver.
Answer:
Given data
d = 10.5 g/cm³
a = 4.07 × 10^-8 cm
Z = 4 atoms
N_A = 6.023 × 10²³
Formula M = \(\frac{\mathrm{d} \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{Z}}\)
= \(\frac{10.5 \times\left(4.07 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{4}\)
= \(\frac{10.5 \cdot \times 67.767 \times 10^{-24} \times 6.023 \times 10^{23}}{4}\)
= 107.09 gm/mole.

Question 3.
Niobium crystallizes in body – centered cubic structure. If density is 8.55 g cm^-3, calculate atomic radius of niobium using its atomic mass 93 U.
Answer:
Radius of unit cell in bcc structure = \(\frac{\sqrt{3}}{4}\) a
First we have to calculate edge length of unit cell ‘a’
Given Atomic mass of Niobium = 93 g/mole
No. of particles in bcc type unit cell (Z) = 2
mass of unit cell = \(\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}}}=\frac{2 \times 93}{6.023 \times 10^{23}}\) = 30.89 × 10²³ gms
Given Density (d) = 8.55 gm/cm³
Volume of unit cell (a³) = \(\frac{\text { mass }}{\text { density }}=\frac{30.89 \times 10^{23}}{8.55}\)
= 36.16 × 10^-24 cm³.
Edge length of unit cell (a) = (36.13 × 10^-24)^1/3
= 3.31 × 10^-8 cm
radius of Unit cell (r) = \(\frac{\sqrt{3}}{4}\) a
= \(\frac{\sqrt{3} \times 3.31 \times 10^{-8}}{4}\)
= 1.43 × 10^-8 cm = 143 pm.

Question 4.
Copper crystallizes into a FCC lattice with edge length 3.61 × 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g.cm^-3.
Answer:
Density d = \(\frac{\mathrm{ZM}}{\mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}\)
Given edge length 3.61 × 10^-8 cm
For FCC lattice of copper, Z = 4
Atomic mass of copper M = 63.5 gms/mole
d = \(\frac{4 \times 63.5}{\left(3.61 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}\)
= 8.97 g/cm³
The calculated value is approximately in agreement with the measured value 8.92 g/cm³.

Question 5.
Ferric oxide crystallizes in a hexagonal close – packed array of oxide ions with two of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
Answer:
Given ferric oxide crystallises in a hexagonal close packed array of oxide ions with two out of every three octahedral,holes occupied by ferric ions.
In hexagonal close – packed arrangement there is one octahedral hole for each atom. If the number of oxide ions (O^-2) per unit cell is one, then the number of Fe⁺³ ions = 2/ 3 × octahedral holes.
= 2/3 × 1 = 2/3
The formula of the compound = Fe_2/3 O₁ (or) Fe₂ O₃.

Question 6.
Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm.
i) What is the length of the side of the unit cell.
ii) how many unit cells are there in 1.00 cm³ of aluminium.
Answer:
i) Given radius = 125 pm
For a fcc lattice unit cell r = \(\frac{a}{2 \sqrt{2}}\)
a = 2\(\sqrt{2}\) × r
= 2 × 1.414 × 125 = 353.5 pm
∴ Length of the side of the unit cell = 353.5 pm

ii) Volume of unit cell = a³ = (353.5 × 10^-10 cm)³
= 442 × 10^-25 cm³
Number of unit cell = \(\frac{1}{4.42 \times 10^{-25}}\)
= 2.26 × 10²² unit cells.

Question 7.
How do you obtain the diffraction pattern for a crystalline substance ?
Answer:
Diffraction of electromagnetic radiation takes place when a beam of light is scattered by an object containing regularly spaced lines (or) points. This scattering phenomenon can happen only if the spacing between the lines (of) points is comparable to the wave length of the radiation.

From the following figure diffraction is due to interface between two waves passing through the same region of space at the same time.
If the waves are in phase, peak to peak and trough to trough, the interference is constructive and the combined wave is increased intensity.

• If the waves are out of phase, the interference is destructive interference and the wave is cancelled.
• Constructive interference gives rise to intense spots observed on Laue’s photographic plate, while destructive interference causes the surrounding high) areas.
  

Derivation of Bragg’s equation : When X-rays are incident on the crystal or plane, they are diffracted from the lattice points (lattice points may be atoms or ions or molecules). In the crystal the lattice points are arranged in regular pattern. When the waves are diffracted from these points, the waves may be constructive or destructive interference.

The 1^st and 2^nd waves reach the crystal surface. They undergo constructive interference. Then from the figure 1 and 2h1 rays are parallel waves. So, they travel the same distance till the wave front AD. The second ray travels more than the first by an extra distance (DB + BC) after crossing the grating for it to interfere with the first ray in a constructive manner. Then only they can be in the same phase with one another. If the two waves are to be in phase, the path difference between the two ways must be equal to the wavelength (λ) or integral multiple of it (nλ, where n = 1, 2, 3 )
(i.e.,) nλ = (DB + BC) [where n = order of diffraction]
DB = BC = d sin θ [θ = angle of incident beam,]
(DB + BC) = 2d sin θ [d = distance between the planes]
nλ = 2d sin θ
This relation is known as Bragg’s equation.

Textual Examples

Question 1.
A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make ccp and those of the element X (as cations) occupy all the octahedral voids. What is the formula of the compound ?
Solution:
The ccp lattice is formed by the element Y. The number of octahedral voids generated would be equal to the number of atoms of Y present in it. Since all the octahedral voids are occupied by the atoms of X, their number would also be equal to that of the element Y. Thus, the atoms of elements X and Y are present in equal numbers or 1 : 1 ratio. Therefore, the formula of the compound is XY.

Question 2.
Atoms of element B form hep lattice and those of the element A occupy 2/3^rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B ?
Solution:
The number of tetrahedral voids formed is equal to twice the number of atoms of element B and only 2/3^rd of these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is 2 × (2/3) : 1 or 4 : 3 and the formula of the compound is A₄B₃.

Question 3.
An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm³. How many atoms are present in 208 g of the element ?
Solution:
Volume of the unit cell = (288 pm)³
= (288 × 10^-12 m)³ = (288 × 10^-10 cm)³ = 2.39 × 10^-23 cm³.
Volume of 208 g of the element
= \(\frac{\text { mass }}{\text { density }}=\frac{208 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}\) = 28.88 cm³
Number of unit cells in this volume
= \(\frac{28.88 \mathrm{~cm}^{3}}{2.39 \times 10^{-23} \mathrm{~cm}^{3} / \text { Unit cell }}\) = 12.08 × 10²³ unit cells
Since each bcc cubic unit cell contains 2 atoms, therefore, the total number of atoms in 208 g = 2 (atoms/unit cell) × 12.08 × 10²³ unit cells
= 24.16 × 10²³ atoms

Question 4.
X-ray diffraction studies show that copper crystallises in ah fee unit cell with cell edge of 3.608 x 10-8 cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm3, calculate the atomic mass of copper.
Solution:
In case of fee lattice, number of atoms per unit cell, z = 4 atoms.
Therefore, M = \(\frac{\mathrm{dN}_{\mathrm{A}} \mathbf{a}^{3}}{z}\)
= \(\frac{8.92 \mathrm{~g} \mathrm{~cm}^{-3} \times 6.022 \times 10^{23} \text { atoms mol}^{-1} \times\left(3.608 \times 10^{-8} \mathrm{~cm}\right)^{3}}{4 \text { atoms }}\) = 63.1 g/mol
Atomic mass of copper = 63.1 u

Question 5.
Silver forms ccp lattice and X-ray studies of its ciystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u).
Solution:
Since the lattice is ccp, the number of silver atoms per unit celll = z = 4
Molar mass of silver = 107.9 g mol^-1 = 107.9 × 10^-3 kg mol^-1.
Edge length of unit cell = a = 408.6 pm = 408.6 × 10^-12 m
Density, d = \(\frac{\mathrm{z} \cdot \mathrm{M}}{\mathrm{a}^{3} \cdot \mathrm{N}_{\mathrm{A}}}\)
= \(\frac{4 \times\left(107.9 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}{\left(408.6 \times 10^{-12} \mathrm{~m}\right)^{3}\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}\)
= 10.5 × 10³ kg m^-3
= 10.5 g cm³.

Intext Questions

Question 1.
Why are solids rigid ?
Answer:
In solid state the constituent particles are not free to move. They can only oscillate about their mean positions due to strong attraction forces between the particles. That is why solids have a closely packed arrangement and rigid structure.

Question 2.
Why do solids have a definite volume ?
Answer:
The constituents particles in solids are bound to th.eir mean positions by strong forces of attraction. The interparticle distances remain unchanged even at increased or reduced pressure. Therefore, solids have a definite volume.

Question 3.
Classify the following as amorphous or crystalline solids : polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Answer:
Amorphous solids
Polyurethane
Naphthalene
Teflon
Cellophane
Polyvinyl chloride
Fibre glass

Crystalline solids
Benzoic acid
Potassium nitrate
Copper

Question 4.
Why is glass considered a supercooled liquid ?
Answer:
Liquids have the characteristic property i.e., the tendency to flow. Glass also shows this property, though it flows very slowly. Glass panes fixed to windows or doors of old buildings are invariably found to be slightly thicker at the bottom than the top. This is because the glass flows down very slowly and makes the bottom portion slightly thicker. Therefore, glass is considered as a supercooled liquid.

Question 5.
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property ?
Answer:
A solid has same value of refractive index along all directions is isotropic and hence amorphous in nature. It would not show a clean cleavage when cut with a knife. Instead, it would break into pieces With irregular surface.

Question 6.
Classify the following solids in diff rent categories based on the nature of intermolecular forces operating in them :
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer:
Ionic solids : Potassium sulphate, zinc sulphide (as they have iohic bonds).
Covalent solids : Graphite, silicon carbide (as they are covalent giant molecules).
Molecular solids : Benzene, urea, ammonia, water, argon (as they have covalent bond).
Metallic solids : Rubidium, tin (as these are metals).

Question 7.
Solid A’ is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it ?
Answer:
Since, the solid ‘A’ is an insulator in solid as well as in molten state, it shows the absence of ions in it. Moreover it melts at extremly high temperature, so it is a giant molecule. These are the properties of covalent solids. So, it is a covalent solid.

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In ionic solids, electrical conductivity is due to the movement of ions. In solid state, ions cannot move and remain held together by strong electrostatic forces of attraction. Therefore, they behave as insulators.

Question 9.
What type of solids are electrical conductors, malleable and ductile ?
Answer:
Metallic solids.

Question 10.
Ire the significance of a ‘lattice point’.
Answer:
The lattice point denotes the position of a particular constituent (atom, ion or molecule) in a crystal lattice. The arrangement of the lattice points in space is responsible for the shape of a particular crystalline solid.

Question 11.
Name the parameters that characterise a unit cell.
Answer:
A unit cell is characterised by

1. its dimensions along the three edges, a, b and c.
2. angles between the edges, α (between b and c) β ( between a and c) and γ (between a and b).
   Thus, a unit cell is characterised by six parameters, a, b, c, α, β and γ.

Question 12.
Distinguish between
i) hexagonal and monoclinic unit cells.
ii) face-centered and end-centered unit cells.
Answer:

Question 13.
Explain how much portion of an atom located at

1. Corner ?
2. body-centre of a cubic unit cell is part of its neighbouring unit cell ?

Answer:

1. A point lying at the corner of a unit cell is shared equally by eight unit cells and therefore,
2. A body centred point belongs entirely to one unit cell since, it is not shared by any other unit cell.

Question 14.
What is the two dimensional coordination number of a molecule in square close packed layer ?
Answer:
Four (4),.as each atom is surrounded by four other atoms.

Question 15.
A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it ? How many of these are tetrahedral voids ?
Answer:
Total number of atoms (N) in a closed packed structure (0.5 mol)
= 0.5 × 6.022 × 10²³ = 3.011 × 10²³
Number of octahedral voids = N = 3.011 × 10²³
Number of tetrahedral voids = 2N = 2 × 3.011 × 10²³ = 6.022 × 10²³
Total number of voids = 3.011 × 10²³ + 6.022 × 10²³
= 9.033 × 10²³

Question 16.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy \(\frac{1}{3}\) rd of tetrahedral voids. What is the formula of the compound ?
i) Find the number of tetrahedral voids as number of tetrahedral voids = 2 × number of atoms present in the lattice.
ii) Calculate the number of atoms (or ratio) of elements M and N as a chemical formula represents the number of atoms of different elements presents in a compound.
iii) Derive the’ formula.
Answer:
Suppose atoms of element N present in ccp = x
Then number of tetrahedral voids = 2x
Since, \(\frac{1}{3}\) rd of the tetrahedral voids are occupied by atoms of elements M.
Therefore, number of atoms of element M = \(\frac{1}{3}\) × 2x = \(\frac{2x}{3}\)
Ratio of M : N = \(\frac{2x}{3}\) : x = 2 : 3
Hence, formula of the compound = M₂N₃.

Question 17.
Which of the following lattices has the highest packing efficiency ?

1. Simple cubic
2. Body-centred cubic
3. Hexagonal close-packed lattice

Packing efficiency in .

1. Simple cubic lattice = 52.4%
2. body-centred cubic lattice = 68%
3. hexagonal close-packed lattice = 74%

Answer:
Hexagonal closed packed lattice has the highest packing efficiency (74%)

Question 18.
An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 10³ kg m^-3 what is the nature of the cubic unit cell ?
Answer:
Density (d) = \(\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}\) so, Z = \(\frac{\mathrm{d} \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{M}}\)
Given, M = 2.7 × 10^-2 kg mol^-1
a = 405 pm = 405 × 10^-12 m = 4.05 × 10^-10m
d = 2.7 × 10³ kgm^-3
NA = 6.022 × 10²³ mol^-1.
Hence.
Z = \(\frac{\left(2.7 \times 10^{3 .} \mathrm{kg} \mathrm{m}^{-3}\right)\left(4.05 \times 10^{-10} \mathrm{~m}\right)^{3} \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}{\left(2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}\right)}\)
= 3.99 = 4
Since, there are four atoms per unit cell, the cubic unit cell must be face – centred.

Question 19.
What type of defect can arise when a solid is heated ? physical property is affected by it and in what way ?
Answer:
When a solid is heated, a vacancy is created in the crystal. On heating, some of the lattice sites are vacant and the density of the solid decreases as the number of ions per unit volume decreases.

Question 20.
What type of stoichiometric defect is shown by

1. ZnS
2. AgBr

Answer:

1. ZnS shows Frenkel defect because its ions have large difference in size.
2. AgBr shows both Frenkel and Schottky defects.

Question 21.
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is aded as an impurity in it ?
Answer:
When a cation of higher valence is added as an impurity to an ionic solid, some vacancies are created. This can be explained with the help of an example. When strontium, chloride (SrCl₂) is added as an impurity to ionic solid sodium chloride (NaCl), two vacant sites are created by removal of one Na⁺ ion. One vacant site is replaced by Sr²⁺ ion but the other remains vacant. The reason is that the crystal as a whole is to remain electrically neutral. Note : [Cationic vacancies produced = No. of cations of higher valency × difference in ’ valencies of the original cation and cation of higher valency]

Question 22.
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Answer:
We can explain the metal excess defect with the example of sodium chloride crystals. When NaCl crystals are heated in an atmosphere of sodium vapour, the Na atoms are deposited on the surface of the crystal. The Cl^– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCZ. This happens by loss of electron by Na atoms to form Na⁺ ions. The released electrons occupy anionic sites by diffusing into the crystals. These electrons absorb energy from visible light and emit radiations corresponding to yellow colour. These electrons are called F-centres (from the German word Farbenzenter meaning colour centre).

Question 23.
A group 14 elements is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong ?
Answer:
n-type semiconductor means increase in conductivity due to presence of excess of electrons. Therefore, a 14 group element should be doped with a 15-group element Eg : arsenic or phosphorus.

Question 24.
What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic ? Justify your answer
Answer:
Ferromagnetic substances make better permanent magnets than ferrimagnetic substances. The metal ions of a ferromagenetic substance are grouped into small regions known as domains and these are randomly oriented. When a magnetic field is applied, all domains are oriented in the direction of the magnetic field. Now the ferromagnetic substance behaves as a magnet. When the applied magnetic field is removed, the magnetic character is retained. Thus, the ferromagnetic substance becomes a permanent magnet.
This property (of being permanently magnetised) is not found in ferrimagnetic substances. They lose their magnetic property on heating.

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Parabola Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Parabola Important Questions

Question 1.
Find the equation of the parabola whose focus is S (1, -7) and vertex is A(1, -2). [T.S. Mar. 15]
Solution:
Let S = (1, -7), A(1, -2)
h = 1, k = -2, a = -2 + 7 = 5
Axis of the parabola is parallel to y – axis
Equation of the parabola is
(x – h)² = – 4a (y – k)
(x – 1)² = – 20(y + 2)
x² – 2x + 1, = – 20y – 40
⇒ x² – 2x + 20y + 41 – 0.

Question 2.
If the normal at the point t₁ on the parabola y² =, 4ax meets it again at point t² then prove that t₁t₂ + t₁² + 2 = 0. [May 07]
Solution:
Equation of normal is
y – y₁ = \(\frac{-y_{1}}{2 a}\) (x – x₁)
y – 2at₁ = \(\frac{-2 a t_{1}}{2 a}\) (x – at₁²) ……………… (i)
Equation of the line (i) again meets parabola at (at₂², 2at₁)
∴ 2at₂ – 2at₁ = t₁ (at₂² – at₁²)
\(-\frac{2}{t_{1}}\) = t₁ + t₂ ⇒ -2 = t₁² + t₁t₂
⇒ t₁² + t₁t₂ + 2 = 0

Question 3.
Find the coordinates of the vertex and focus, and the equations of the directrix and axes of the y – x + 4y + 5 = 0. [Mar 05]
Solution:
y² – x + 4y + 5 = 0 ⇒ (y – (-2))² = (x – 1), comparing with (y – k)² = 4a(x – h),we get (h, k) = (1, -2) and a = \(\frac{1}{4}\), coordinates of the vertex (h, k) = (1 ,-2)
coordinates of the focus (h + a, k) = (\(\frac{5}{4}\), -2)
Equation of the directrix x – h + a = 0
i.e., 4x – 3 = 0
Equation of the axis y – k = 0.
i.e., y + 2 = 0

Question 4.
Find the coordinates of the points on the parabola y² = 2x whose focal distance is \(\frac{5}{2}\). [A.P. Mar. 15, Mar. 13]
Solution:
Let P(x₁, y₁) be a point on the parabola
y² = 2x whose focal distance is \(\frac{5}{2}\) then
y₁² = 2x₁ and x₁ + a = \(\frac{5}{2}\)
⇒ x₁ + \(\frac{1}{2}\) = \(\frac{5}{2}\) ⇒ x₁ = 2
∴ y₁² = 2(2) = 4 ⇒ y₁ = ±2
∴ The required points are (2, 2) and (2, -2)

Question 5.
Find the value of k if the line 2y = 5x + k is a tangent to the parabola y² = 6x. [T.S. Mar. 16]
Solution:
Given line is 2y = 5x + k
⇒ y = (\(\frac{5}{2}\))x + (\(\frac{k}{2}\))
Comparing y = (\(\frac{5}{2}\))x + (\(\frac{k}{2}\)) with y = mx + c
We get m = \(\frac{5}{2}\), c = \(\frac{k}{2}\)
y = (\(\frac{5}{2}\))x + \(\frac{k}{2}\) is a tangent to y² = 6x
c = \(\frac{a}{m}\)
⇒ \(\frac{k}{2}=\frac{\left(\frac{3}{2}\right)}{\left(\frac{5}{2}\right)}\) ⇒ k = \(\frac{6}{5}\)

Question 6.
Show that the equations of common tangents to the circle x² + y² = 2a² and the parabola y² = 8ax are y = ± (x + 2a). [Mar. 06]
Solution:
The equation of tangent to parabola
y² = 8ax is y = mx + \(\frac{2a}{m}\)
m²x – my + 2a = 0 ……………….. (1)
If (i) touches circle x² + y² = 2a², then the length of perpendicular from its centre (0, 0) to (i) must be equal to the radius a\(\sqrt{2}\) of the circle.
\(\left|\frac{2 a}{\sqrt{m^{2}+m^{4}}}\right|=a \sqrt{2}\)
or 4 = 2 (m⁴ + m²)
m⁴ + m² – 2 = 0
(m² + 2) (m² – 1) = 0 or m = ±1
Required tangents are
y = (1) x + \(\frac{2 a}{(1)}\), y = (-1) x + \(\frac{2 a}{(-1)}\)
y = ± (x + 2a)

Question 7.
Find the co-ordinates of the point on the parabola y² = 8x whose focal distance is 10. [T.S. Mar. 17] [A.P. Mar. 16; Mar. 14, 11]
Solution:
Equation of the parabola is
y² = 8x
4a = 8 ⇒ a – 2

Co-ordinates of the focus S are (2, 0) Suppose P(x, y) is the point on the parabola.
Given SP = 10 ⇒ SP² = 100
(x – 2)² + y² = 100
But y² = 8x
⇒ (x – 2)² + 8x = 100
⇒ x² – 4x + 4 + 8x – 100 = 0
⇒ x² + 4x – 96 = 0 ⇒ (x + 12) (x – 8) = 0
x + 12 = 0 or x – 8 = 0
x = -12, or 8
Case (i) x = 8
y² = 8.x = 8.8 = 64
y = ±8
Co-ordinates of the required points are (8, 8) and (8, -8)
Case (ii) x = -12
y² = 8(-12) = -96 < 0
y is not real.

Question 8.
If (\(\frac{1}{2}\), 2) is one extermity of a focal chord of the parabola y² = 8x. Find the co-ordinates of the other extremity. [May 06]
Solution:
A = (\(\frac{1}{2}\), 2); S = (2, 0)
B = (x₁, y₁) ⇒ (\(\frac{y_{1}^{2}}{8}\), y₁)
ASB is a focal chord.
∴ Slopes of SA and BS are same.

or 4y₁² + 24y₁ – 64 = 0
⇒ y₁² + 6y₁ – 16 = 0
⇒ (y₁ + 8) (y, – 2) = 0
y₁ = 2, 8
x₁ = \(\frac{1}{2}\), 8; So (8, -8) other extremity.

Question 9.
Show that the common tangent to the parabola y² = 4ax and x² = 4by is xa^1/3 + yb^1/3 + a^2/3b^2/3 = 0 [T.S. Mar. 17] [A.P. Mar. 16]
Solution:
The equations of the parabolas are
y² = 4ax ………………. (1)
and x² = 4by ……………… (2)
Equation of any tangent to (1) is of the form
y = mx + \(\frac{a}{m}\) …………….. (3)
If the line (3) is a tangent to (2) also, we must get only one point of intersection of (2) and (3).
Substituting the value of y from (3) in (2), we get x² = 4b (mx + \(\frac{a}{m}\)) is 3x² – 4bm²x – 4ab = 0 should have equal roots Therefore its discriminent must be zero. Hence
16b²m⁴ – 4m (-4ab) = 0
16b(bm⁴ + am) = 0
m(bm³ + a) = 0 But m ≠ 0
∴ m = – a^1/3/b^1/3 substituting in (3) the equation of the common tangent becomes
y = \(-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}\) or
a^1/3x + b^1/3y + a^2/3b^2/3 = 0 .

Question 10.
Prove that the area of the triangle formed by the .tangents at (x₁, y₁), (x₂, y₂) and (x₃, y₃) to the parabola y² = 4ax (a > 0) is \(\frac{1}{16a}\)|(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)| sq. units. [T.S. Mar. 15]
Solution:
Let D(x₁, y₁) = (at₁², 2at₁),
E(x₂, y₂) = (at₂², 2at₂),
and F(x₃, y₃) = (at₃², 2at₃), be three points on the parabola
y² = 4ax (a > 0).
The equation of the tangents at D, E and F are
t₁y = x + at₁² ……………… (1)
t₂y = x + at₂² ………………. (2)
t₃y = x.+ at₃² ………………. (3)
(1) – (2) ⇒ (t₁ – t₂) y = a(t₁ – t₂) (t₁ + t₂)
⇒ y = a(t₁ + t₂) substituting in (1)
we get x = at₁t₂.
∴ The point of intersection of the tangents at D and E is say P(at₁t₂, a(t₁ + t₂))
Similarly the points of intersection of tangent at E, F and at F, D are Q(at₂t₃, a(t₂ + t₃) and R(at₃t₁ a(t₃ + t₁)) respectively
Area of ∆PQR

\(\frac{1}{16a}\) |2a(t₁ – t₂) 2a(t₂ – t₃) 2a(t₃ – t₁)|
\(\frac{1}{16a}\)|(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)| sq. units.

Question 11.
Prove that the two parabolas y² = 4ax and x² = 4by intersect (other than the origin) at an angle of Tan^-1 \(\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\) [Mar. 14]
Solution:
Without loss of generality we assume a > 0 and b > 0.
Let P(x, y) be the point of intersection of the parabolas other than the origin. Then
y⁴ = 16a²x²
= 16a²(4by)
= 64a²by
∴ y[y³ – 64a²b] = 0
=> y³ – 64a²b = 0
=> y = (64a²b)^1/3 [∵ y > 0]
= 4a^2/3b^1/3

Also from y² = 4ax, x = \(\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}\)
= 4a^1/3b^2/3
∴ P = (4a^1/3b^2/3, 4a^2/3b^1/3)
Differentiating both sides of y² 4ax w.r.t ‘x’, we get
\(\frac{d y}{d x}=\frac{2 a}{y}\)
∴ \(\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}\)
If m₁ be the slope of the tangent at P to y² = 4ax, then
m₁ = \(\frac{1}{2}\left(\frac{a}{b}\right)^{1 / 3}\)
Similarly, we get m₂ = \(2\left(\frac{a}{b}\right)^{1 / 3}\) where m₂ is the slope of the tangent at P to x² = 4by.
If θ is the acute angle between the tangents to the curves at P, then
tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\)
so that θ = \(\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\)

Question 12.
Find the coordinates of the vertex and focus, and the equations of the directrix and axes of the following parabolas.
(i) y² = 16x
(ii) x² = -4y
(iii) 3x² – 9x + 5y – 2 = 0
(iv) y² – x + 4y + 5 = 0 [Mar. 05]
Solution:
i) y² = 16x, comparing with y² = 4ax,
we get 4a = 16 ⇒ a = 4
The coordinates of the vertex = (0, 0)
The coordinates of the focus = (a, 0) = (4, 0)
Equation of the directrix: x + a = i.e., x + 4 = 0
Axis of the parabola y = 0

ii) x² = -4y, comparing with x² = -4ay,
we get 4a = 4 ⇒ a = 1
The coordinates of the vertex = (0, 0
The coordinates of the focus = (0, -a) = (0, 1)
The equation of the directrix y – a = 0
i.e., y – 1 = 0 .
Equation of the axis x = 0

iii) 3x² – 9x + 5y – 2 = 0
3(x² – 3x) = 2 – 5y
⇒ 3(x² – 2x(\(\frac{3}{2}\)) + \(\frac{9}{4}\)) = 2 – 5y + \(\frac{27}{4}\)
(x – \(\frac{3}{2}\))² = –\(\frac{5}{3}\) (y – \(\frac{7}{4}\)),
Comparing with (x – h)² = -4a (y – k) we get
a = \(\frac{5}{12}\), h = \(\frac{3}{2}\), k = \(\frac{7}{4}\)
∴ Coordinates of the vertex = (h, k)
= (\(\frac{3}{2}\), \(\frac{7}{4 }\))
Coordinates of the focus = (h, k – a)
= (\(\frac{3}{2}\), \(\frac{7}{4}\) – \(\frac{5}{12}\)) = (\(\frac{3}{2}\), \(\frac{4}{3}\))
Equation of the directrix is y – k – a = 0
i.e., 6y – 13 = 0
Equation of the axis is x – h = 0
i.e., 2x – 3 = 0

iv) y² – x + 4y + 5 = 0 ⇒ (y- (-2))² = (x – 1),
comparing with (y – k)² = 4a(x – h),we get
(h, k) = (1, -2) and a = \(\frac{1}{4}\), coordinates of the vertex (h, k) = (1 ,-2)
coordinates of the focus
(h + a, k) = (\(\frac{5}{4}\), -2)
Equation of the directrix x – h + a = 0
i.e., 4x – 3 = 0
Equation of the axis y – k = 0.
i.e., y + 2 = 0

Question 13.
Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).
Solution:
The abcissae of the vertex and focus are equal to 3. Hence the axis of the parabola is x = 3, a line parallel to y-axis, focus is above the vertex.
a = distance between focus and vertex = 3.
∴ Equation of the parabola
(x – 3)² = 4(3) (y + 2)
i.e., (x – 3)² = 12(y + 2).

Question 14.
Find the coordinates of the points on the parabola y² = 2x whose focal distance is \(\frac{5}{2}\). [A.P. Mar. 15, Mar. 13]
Solution:
Let P(x₁, y₁) be a point on the parabola
y² = 2x whose focal distance is \(\frac{5}{2}\) then
y₁² = 2x₁ and x₁ + a = \(\frac{5}{2}\)
⇒ x₁ + \(\frac{1}{2}\) = \(\frac{5}{2}\) ⇒ x₁ = 2
∴ y₁² = 2(2) = 4 ⇒ y₁ = ±2
∴ The required points are (2, 2) and (2, -2)

Question 15.
Find the equation of the parabola passing through the points (-1, 2), (1, -1) and (2, 1) and having its axis parallel to the X-axis.
Solution:
Since the axis is parallel to x-axis the equation of the parabola is in the form of
x = -ly² + my + n. .
Since the parabola passes through (-1, 2), we have
-1 = l( 2)² + m(2) + n
⇒ 4l + 2m + n = – 1 ……………………. (1)
Similarly, since the parabola passes through (1,-1) and (2, 1) we have
l – m + n = 1 ……………… (2)
l + m + n = 2 ……………… (3)
Solving (1), (2) and (3)
we get l = –\(\frac{7}{6}\), m = \(\frac{1}{2}\) and n = \(\frac{8}{3}\).
Hence the equation of the parabola is
x = –\(\frac{7}{6}\) y² + \(\frac{1}{2}\) y + \(\frac{8}{3}\) (or)
7y² – 3y + 6x- 16 = 0.

Question 16.
A double ordinate of the curve y² = 4ax is of length 8a. Prove that the line from the vertex to its ends are at right angles.
Solution:
Let P = (at², 2at) and P’ = (at², -2at) be the ends of double ordinate PP’. Then
8a = PP’ = \(\sqrt{0+(4 a t)^{2}}\) = 4at ⇒ t = 2.
∴ P = (4a, 4a), P’= (4a, -4a) .
Slope of \(\overline{\mathrm{AP}}\) × slope of \(\overline{\mathrm{AP}^{\prime}}\)
= (\(\frac{4a}{4a}\))(-\(\frac{4a}{4a}\)) = -1
∴ ∠PAP’ = \(\frac{\pi}{2}\)

Question 17.
i) If the coordinates of the ends of a focal chord of the parabola y² = 4ax are (x₁, y₁) and (x₂, y₂), then prove that x₁x₂ = a², y₁y₂ = -4a².
ii) For a focal chord PQ of the parabola y² = 4ax, if SP = l and SQ = l’ then prove that \(\frac{1}{l}\) + \(\frac{1}{l}\) = \(\frac{1}{a}\).
Solution:
i) Let P(x₁, y₁) = (at₁², 2at₁) and Q(x₂, y₂) = (at₂², 2at₁) be two end points of a focal chord.
P, S, Q are collinear
Slope of \(\overline{\mathrm{PS}}\) = Slope of \(\overline{\mathrm{QS}}\)
\(\frac{2 a t_{1}}{a t_{1}^{2}-a}=\frac{2 a t_{2}}{a t_{2}^{2}-a}\)
t₁t₂² – t₁ = t₂t₁² – t₂
t₁t₂ (t₂ – t₁) + (t₂ – t₁) = 0
1 + t₁t₂ = 0 ⇒ t₁t₂ = -1 ………………… (1)
From (1) x₁x₂ = at₁² at₂² = a²(t₂t₁)² = a²
y₁y₂ = 2at₁2at₂ = 4a²(t₂ t₁) = -4a²

ii) Let P(at₁², 2at₁) and Q(at₂², 2at₁) be the extremities of a focal chord of the parabola, then t₁t₂ = -1 (from (1))

Question 18.
If Q is the foot of the perpendicular from a point P on the parabola y² = 8(x – 3) to its directrix. S is the focus of the parabola and if SPQ is an equilateral triangle then find the length of side of the triangle.
Solution:
Given parabola y² = 8(x – 3) then
its vertex A = (3, 0) and focus = (5, 0)
[4a = 8 ⇒ a = 2] since PQS is an equilateral triangle

Hence length of each side of triangle is ‘8’.

Question 19.
The cable of a uniformely loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 72 mt. long is supported by vertical wires attached to the cable, the longest being 30 mts. and the shortest being 6 mts. Find the length of the supporting wire attached to the road-way 18 mts. from the middle.

Solution:
Let AOB be the cable [0 is its lowest point and A, B are the highest points]. Let PRQ be the bridge suspended with PR = RQ = 36 mts. (see above Fig.).

PA = QB = 30 mts (longest vertical sup-porting wires)

OR = 6 mts (shortest vertical supporting wire) [the lowest point of the cable is up¬right the mid-point R of the bridge]

Therefore, PR = RQ = 36 mts. We take the origin of coordinates at 0, X-axis along the tangent at O to the cable and the Y-axis along \(\overleftrightarrow{\mathrm{RO}}\). The equation of the cable would, therefore, be x² = 4ay for some a > 0. We get B = (36, 24) and 36² = 4a × 24.
Therefore, 4a = \(\frac{36 \times 36}{24}\) = 54 mts.
If RS = 18 mts. and SC is the vertical through S meeting the cable at C and the X-axis at D, then SC is the length of the supporting wire required. If SC = l mts, then
DC = (l – 6) mts. As such C = (18, l – 6).
Since C is on the cable, 18² = 4a (l – 6)
⇒ l – 6 = \(\frac{18^{2}}{4 a}\) = \(\frac{18 \times 18}{54}\) = 6
⇒ l = 12

Question 20.
Find the condition for the straight line lx + my + n = 0 to be a tangent to the parabola y² = 4ax and find the co-ordinates of the point of contact.
Solution:
Let the line lx + my + n = 0 be a tangent to the parabola y² = 4ax at (at², 2at). Then the equation of the tangent at P(t) is x – yt + at² = 0 then it represents the given line
lx + my + n = 0, then

Question 21.
Show that the straight line 7x + 6y = 13 is a tangent to the parabola y² – 7x – 8y + 14 = 0 and find the point of contact.
Solution:
Equation of the given line is 7x + 6y = 13, equation of the given parabola is y² – 7x – 8y + 14 = 0.
By eliminating x, we get the ordinates of the points of intersection of line and parabola adding the equations y² – 2y + 1 = 0.
i.e., (y – 1 )² = 0 ⇒ y = 1, 1.
∴ The given line is tangent to the given parabola.
If y = 1 then x = 1 hence the point of contact is (1, 1).

Question 22.
Prove that the normal chord at the point other than origin whose ordinate is equal to its abscissa subtends a right angle at the focus.
Solution:
Let the equation of the parabola be
y² = 4ax and P(at², 2at) be any point …………………. (1)
On the parabola for which the abscissa is equal to the ordinate.
i.e., at² = 2at ⇒ t = 0
or t = 2. But t ≠ 0. Hence the point (4a, 4a) at which the normal is
y + 2x = 2a(2) + a(2)³ (or)
y = (12a – 2x) ………………….. (2)
Substituting the value of
y = 12a – 2x in (1) we get
(12a – 2x)² = 4ax (or)
x² – 13ax + 36a² = (x – 4a) (x – 9a) = 0
⇒ x = 4a, 9a
corresponding values of y are 4a and -6a. Hence the other points of intersection of that normal at P(4a, 4a) to the given parabola is Q(9a, -6a), we have S(a, 0).
Slope of the \(\overline{\mathrm{SP}}\) = m₁ = \(\frac{4a-0}{4a-a}\) = \(\frac{4}{3}\),
Slope of the \(\overline{\mathrm{SQ}}\) = m₂ = \(\frac{-6a-0}{9a-a}\) = \(\frac{3}{4}\)
clearly m₁m₂ = -1, so that \(\overline{\mathrm{SP}}\) ⊥ \(\overline{\mathrm{SQ}}\)

Question 23.
From an external point P, tangent are drawn to the parabola y² = 4ax and these tangent make angles θ₁, θ₂ with its axis, such that tan θ₁ + tan θ₂ is constant b. then show that P lies on the line y = bx.
Solution:
Let the coordinates of P be (x₁, y₁) and the equation of the parabola y² = 4ax. Any tangent to .the parabola is y = mx + \(\frac{a}{m}\), if this passes through (x₁, y₁) then
y₁ = mx, + \(\frac{a}{m}\)
i.e., m²x₁ – my₁ + a = 0 ………………… (1)
Let the roots of (1) be m₁, m₂₁.
Then m₁ + m₂ = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\)
⇒ tan θ₁ + tan θ₂ = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\)
[∵ The tangents make angles θ₁, θ₂ with its axis (x – axis) then their slopes m₁ = tan θ₁ and m₂ = tan θ₂].
∴ b = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\) ⇒ y₁ = bx₁
∴ P(x₁, y₁) lies on the line y = bx.

Question 24.
Show that the common tangent to the parabola y² = 4ax and x² = 4by is xa^1/3 + yb^1/3 + a^2/3b^2/3 = 0. [A.P. Mar. 16]
Solution:
The equations of the parabolas are
y² = 4ax ………………. (1)
and x² = 4by …………………. (2)
Equation of any tangent to (1) is of the form
y = mx + \(\frac{a}{m}\)
If the line (3) is a tangent to'(2) also, we must get only one point of intersection of (2) and (3).
Substituting the value of y from (3) in (2),
we get x² = 4b (mx + \(\frac{a}{m}\)) is mx² – 4bm²x – 4ab = 0 should have equal roots therefore its discriminent must be zero. Hence
16b²m⁴ – 4m (-4ab) = 0
16b(bm⁴ + am) = 0
m(bm³ + a) = 0 But m ≠ 0
∴ m = – a^1/3/b^1/3 substituting in (3) the equation of the common tangent becomes
y = \(-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}\) (or)
∴ a^1/3 x + b^1/3y + a^2/3b^2/3 = 0

Question 25.
Prove that the area of the triangle formed ‘ by the tangents at (x₁, y₁), (x₂, y₂) and (x₃, y₃) to the parabola y² = 4ax (a > 0) is \(\frac{1}{16a}\)|(y₁ – y₂) (y₂ – y₃) (y₃ – y₁)| sq. units. [T.S. Mar. 15]
Solution:
Let D(x₁, y₁) = (at₁², 2at₁)
E(x₂, y₂) = (at₃², 2at₂)
and F(x₃, y₃) = (at₂³, 2at₃)
be three points on the parabola
y² = 4ax (a > 0).
The equation of the tangents at D, E and F are
t₁y = x + at₁² ………………. (1)
t₂y = x + at₂² ………………. (2)
t₃y = x + at₃² ………………. (3)
(1) – (2) ⇒ (t₁ – t₂) y = a(t₁ – t₂) (t₁ + t₂)
⇒ y = a(t₁ + t₂) substituting in (1)
we get x = at₁t₂
∴ The point of intersection of the tangents at D and E is say P(at₁t₂, a(t₁ + t₂))
Similarly the points of intersection of tangent at E, F and at F, D are Q(at₂t₃, a(t₂ + t₃) and R(at₃t₁, a(t₃ + t₁)) respectively
Area of ∆PQR

Question 26.
Prove that the two parabolas y² = 4ax and x² = 4by intersect (other than the origin) at an angle of Tan^-1 \(\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\) [Mar. 14]
Solution:
Without loss of generality we assume a > 0 and b > 0.
Let P(x, y) be the point of intersection of the parabolas other than the origin. Then
y⁴ = 16a²x²
= 16a²(4by)
= 64a²by
∴ y[y³ – 64a²b] = 0
⇒ y³ – 64a²b = 0
⇒ y = (64a²b)^1/3 [∵ y > 0]
= 4a^2/3b^1/3

Also from y² = 4ax, x = \(\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}\)
= 4a^1/3b^2/3
∴ P = (4a^1/3b^2/3, 4a^2/3b^1/3)
Differentiating both sides of y² 4ax w.r.t ‘x’, we get
\(\frac{d y}{d x}=\frac{2 a}{y}\)
∴ \(\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}\)
If m₁ be the slope of the tangent at P to y² = 4ax, then
m₁ = \(\frac{1}{2}\left(\frac{a}{b}\right)^{1 / 3}\)
Similarly, we get m₂ = \(2\left(\frac{a}{b}\right)^{1 / 3}\) where m₂ is the slope of the tangent at P to x² = 4by.
If θ is the acute angle between the tangents to the curves at P, then
tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\)
so that θ = \(\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\)

Question 27.
Prove that the orthocenter of the triangle formed by any three tangents to a parabola lies on the directrix of the parabola.
Solution:
Let y² = 4ax be the parabola and
A = (at₁², 2at₁),
B = (at₂², 2at₂),
C = (at₃², 2at₃) be any three points on it.
Now we consider the triangle PQR formed by the tangents to the prabola at A, B, C
where P = (at₁t₂, a(t₁ + t₂)),
Q = (at₂t₃, a(t₂ + t₃)) and R = (at₃t₁, a(t₃ + t₁)).
Equation of \(\overleftrightarrow{\mathrm{QR}}\) (i.e., the tangent at C) is x – t₃ y + at₃² = 0.
Therefore, the attitude through P of triangle PQR is
t₃x + y = at₁t₂t₃ + a(t₁ + t₂) ………………….. (1)
Similarly, the attitude through Q is
t₁x + y = at₁t₂t₃ + a(t₂ + t₃) ………………….. (2)
Solving (1) and (2), we get (t₃ – t₁)
x’ = a(t₁ – t₃) i.e., x = – a.
Therefore, the orthocenter of the triangle PQR, with abscissa as -a, lies on the directrix of the parabola.

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Circle Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B Circle Important Questions

Question 1.
If x² + y² + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3), find g, f and its radius. [Mar. 11]
Solution:
Circle is x² + y² + 2gx + 2fy – 12 = 0
C = (-g, -f) C = (2, 3)
∴ g = – 2, f = – 3, c = – 12
Radius = \(\sqrt{g^{2}+f^{2}-c}\)
= \(\sqrt{4+9+12}\) = 5 units

Question 2.
Obtain the parametric equation of x² + y² = 4 [Mar. 14]
Solution:
Equation of the circle is x² + y² = 4
C (0, 0), r = 2
Parametric equations are
x = – g + r cos θ = 2 cos θ
y = – b + r sin θ = 2 sin θ, 0 < θ < 2π

Question 3.
Obtain the parametric equation of (x – 3)² + (y – 4)² = 8² [A.P. Mar. 16; Mar. 11]
Solution:
Equation of the circle is (x – 3)² + (y – 4)² =8²
Centre (3, 4), r = 8
Parametric equations are
x = 3 + 8 cos θ, y = 4 + 8 sin θ, 0 < θ < 2π

Question 4.
Find the power of the point P with respect to the circle S = 0 when
ii) P = (-1,1) and S ≡ x² + y² – 6x + 4y – 12
Solution:
Power of the point = S₁₁
= 1 + 1 + 6 + 4 – 12 = 0

iii) P = (2, 3) and S ≡ x² + y² – 2x + 8y – 23
Power of the point = S₁₁
= 4 + 9 – 4 + 24 – 23 = 10

iv) P = (2, 4) and S ≡ x² + y² – 4x – 6y – 12
Power of the point = 4 + 16 – 8 – 24 – 12
= -24.

Question 5.
If the length of the tangent from (5, 4) to the circle x² + y² + 2ky = 0 is 1 then find k. [Mar. 15; Mar. 01]
Solution:
= \(\sqrt{S_{11}}=\sqrt{(5)^{2}+(4)^{2}+8 k}\)
But length of tangent = 1
∴ 1 = \(\sqrt{25+16+8k}\)
Squaring both sides we get 1 = 41 + 8k
k = – 5 units.

Question 6.
Find the polar of (1, -2) with respect to x² + y² – 10x – 10y + 25 = 0 [T.S. Mar. 15]
Solution:
Equation of the circle is x² + y² – 10x – 10y + 25 = 0
Equation of the polar is S₁ = 0
Polar of P(1,-2) is
x . 1 + y(-2) – 5(x + 1) – 5(y – 2) + 25 = 0
⇒ x – 2y – 5x – 5 – 5y + 10 + 25 = 0
⇒ -4x – 7y + 30 = 0
∴ 4x + 7y – 30 = 0

Question 7.
Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x² + y² = 35. [T.S. Mar. 16]
Solution:
Equation of the circle is
x² + y² = 35
Polar of P(1, 3) is x. 1 + y. 3 = 35
x + 3y = 35
P(1, 3) and Q(2, k) are conjugate points
The polar of P passes through Q
2 + 3k = 35
3k = 33
k = 11

Question 8.
If the circle x² + y² – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]
Solution:
Equation of the circle is
x² + y² – 4x + 6y + a = 0
2g = – 4, 2f = 6, c = a
g = -2, f = 3, c = a
radius = 4 ⇒ \(\sqrt{g^{2}+f^{2}-c}\) = 4
\(\sqrt{4+9-a}\) = 4
13 – a = 16
a = 13 – 16 = -3

Question 9.
Find the value of ‘a’ if 2x² + ay² – 3x + 2y – 1 =0 represents a circle and also find its radius.
Solution:
General equation of second degree
ax² + 2hxy + by² + 2gx + 2fy + c = 0
Represents a circle, when
a = b, h = 0, g² + f² – c ≥ 0
In 2x² + ay² – 3x + 2y – 1 = 0
a = 2, above equation represents circle.
x² + y² – \(\frac{3}{2}\) x + y – \(\frac{1}{2}\) = 0
2g = – \(\frac{3}{2}\); 2f = 1; c = – \(\frac{1}{2}\)
c = (-g, -f) = (\(\frac{+3}{4}\), \(\frac{-1}{2}\))
Radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}\)
= \(\frac{\sqrt{21}}{4}\) units

Question 10.
If the abscissae of points A, B are the roots of the equation, x² + 2ax – b² = 0 and ordinates of A, B are root of y² + 2py – q² = 0, then find the equation of a circle for which \(\overline{\mathrm{AB}}\) is a diameter. [Mar. 14]
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
x² – x(x₁ + x₂) + x₁x₂ + y² – y (y₁ + y₂) + y₁y₂ = 0
x₁, x₂ are roots of x² + 2ax – b² = 0
y₁, y₂ are roots of y² + 2py – q² = 0
x₁ + x₂ = – 2a
x₁x₂ = -b²

y₁ + y₂ = -2p
y₁y₂ = -q²
Equation of circle be
x² – x (- 2a) – b² + y²– y (- 2p) – q² = 0
x² + 2xa + y² + 2py – b² – q² = 0

Question 11.
Find the equation of a circle which passes through (4, 1) (6, 5) and having the centre on 4x + 3y – 24 = 0 [A.P. Mar. 16; Mar. 14]
Solution:
Equation of circle be x² + y² + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then
4² + 1² + 2g(4) + 2f(1) + c = 0 ………….. (i)
6² + 5² + 2g(6) + 2f(5) + c = 0 ……………… (ii)
Centre lie on 4x + 3y – 24 = 0
∴ 4(-g) + 3 (-f) – 24 = 0
(ii) – (i) we get
44 + 4g + 8f = 0
Solving (iii) and (iv) we get
f = -4, g = -3, c = 15
∴ Required equation of circle is
x² + y² – 6x – 8y + 15 = 0

Question 12.
Find the equation of a circle which is concentric with x² + y² – 6x – 4y – 12 = 0 and passing through (- 2, 14). [Mar. 14]
Solution:
x² + y² – 6x – 4y – 12 = 0 …………… (i)
C = (- g, – f)
= (3, 2)
Equation of circle concentric with (i) be
(x – 3)² + (y – 2)² = r²
Passes through (-2, 14)
∴ (- 2 – 3)² + (14 – 2)² = r²
169 = r²
Required equation of circle be
(x – 3)² + (y – 2)² = 169
x² + y² – 6x – 4y – 156 = 0

Question 13.
Find the equation of the circle whose centre lies on the X-axis and passing through (- 2, 3) and (4, 5). [A.P. & T.S. Mar. 15]
Solution:
x² + y² + 2gx + 2fy + c = 0 ……………… (i)
(- 2, 3) and (4, 5) passes through (i)
4 + 9 – 4g + 6f + c = 0 ………………………. (ii)
16 + 25 + 8g + 10f + c = 0 …………………. (iii)
(iii) – (ii) we get
28 + 12g + 4f = 0
f + 3g = – 7
Centre lies on X – axis then f = 0
g = -, \(\frac{7}{3}\), f = 0, c = \(\frac{67}{3}\) -, we get by substituting g; f in equation (ii)
Required equation will be 3(x² + y²) – 14x – 67 = 0

Question 14.
Find the equation of circle passing through (1, 2); (3, – 4); (5, – 6) three points.
Solution:
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
1 + 4 + 2g + 4f + c = 0 …………………… (i)
9 + 16 + 6g – 8f + c = 0 …………………… (ii)
25 + 36 + 10g – 12f + c = 0 ………………… (iii)
Subtracting (ii) – (i) we get
20 + 4g – 12f = 0
(or) 5 + g – 3f = 0 ……………….. (iv)
Similarly (iii) – (ii) we get
36 + 4g – 4f = 0
(or) 9 + g – f = 0 ………………… (v)
Solving (v) and (iv) we get
f = -2, g = – 11, c = 25
Required equation of circle be x² + y² – 22x – 4y + 25 = 0

Question 15.
Find the length of the chord intercepted by the circle x² + y² – 8x – 2y – 8 = 0 on the line x + y + 1 = 0 [T.S. Mar. 16]
Solution:
Equation of the circle is x² + y² – 8x – 2y – 8 = 0
Centre is C(4, 1), r = \(\sqrt{16+1+8}\) = 5
Equation of the line is x + y + 1 = 0
P = distance from the centre = \(\frac{|4+1+1|}{\sqrt{1+1}}\)
= \(\frac{6}{\sqrt{2}}\) = 3\(\sqrt{2}\)
Length of the chord = 2\(\sqrt{r^{2}-p^{2}}\)
= 2\(\sqrt{25-18}\)
= 2\(\sqrt{7}\) units.

Question 16.
Find the pair of tangents drawn from (1, 3) to the circle x² + y² – 2x + 4y – 11 =0 and also find the angle between them. [T.S. Mar. 16]
Solution:
SS₁₁ = S₁²
(x² + y² – 2x + 4y – 11) (1 + 9 – 2 + 12 – 11) = [x + 3y – 1 (x + 1) + 2 (y + 3) – 11]²
(x² + y² – 2x + 4y – 11) 9 = [5y – 6]²
9x² + 9y² – 18x + 36y – 99
= 25y² + 36 – 60y
9x² – 16y² – 18x + 96y – 135 = 0
cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\) = \(\frac{|9-16|}{\sqrt{(25)^{2}}}\)
= \(\frac{|-7|}{25}\) = \(\frac{7}{25}\) ⇒ θ = cos^-1 (\(\frac{7}{25}\))

Question 17.
Find the value of k if the points (4, 2) and (k, -3) are conjugate points with respect to the circle x² + y² – 5x + 8y + 6 = 0. [T.S. Mar. 17]
Solution:
Equation of the circle is x² + y² – 5x + 8y + 6 = 0
Polar of P(4, 2) is
x . 4 + y . 2 – \(\frac{5}{2}\) (x + 4) + 4 (y + 2) + 6 = 0
8x + 4y – 5x – 20 + 8y + 16 + 12 = 0
3x + 12y + 8 = 0
P(4, 2), Q(k, -3) are conjugate points
Polar of P passes through Q
∴ 3k – 36 + 8 = 0
3k = 28 ⇒ k = \(\frac{28}{3}\)

Question 18.
If (2, 0), (0,1), (4, 5) and (0, c) are concyclic, and then find c. [A.P. & T.S. Mar. 15]
Solution:
x² + y² + 2gx + 2fy + c₁ = 0
Satisfies (2, 0), (0, 1) (4, 5) we get
4 + 0 + 4g + c₁ = 0 …………………. (i)
0 + 1 + 2g. 0 + 2f + c₁ = 0 – (ii)
16 + 25 + 8g + 10f + c₁ = -0 ……………. (iii)
(ii) – (i) we get
– 3 – 4g + 2f = 0
4g – 2f = – 3 ……………… (iv)
(ii) – (iii) we get
– 40 – 8g – 8f = 0 (or)
g + f = – 5 …………………… (v)
Solving(iv) and (v) we get
g = –\(\frac{13}{6}\), f = –\(\frac{17}{6}\)
Substituting g and f values in equation (i) we get
4 + 4 (-\(\frac{13}{6}\)) + c₁ = 0
c₁ = \(\frac{14}{3}\)
Now equation x² + y² – \(\frac{13}{3}\) x – \(\frac{17}{3}\) y + \(\frac{14}{3}\) = 0
Now circle passes through (0, c) then
c² – \(\frac{17}{3}\) c + \(\frac{14}{3}\) = 0
3c² – 17c + 14 = 0
⇒ (3c – 14) (c – 1) = 0
(or)
c = 1 or \(\frac{14}{3}\)

Question 19.
Find the length of the chord intercepted by the circle x² + y² – x + 3y – 22 = 0 on the line y = x – 3. [May 11; Mar. 13]
Solution:
Equation of the circle is
S ≡x² + y² – x + 3y – 22 = 0
Centre C(\(\frac{1}{2}\), –\(\frac{3}{2}\))

Question 20.
Find the equation of the circle with centre (-2, 3) cutting a chord length 2 units on 3x + 4y + 4 = 0 [Mar. 11]
Solution:
Equation of the line is 3x + 4y + 4 = 0
P = Length of the perpendicular .

Length of the chord = 2λ = 2 ⇒ λ = 1
If r is the radius of the circle then
r² = 2² + 1² – 4 + 1 = 5
Centre of the circle is (-2, 3)
Equation of the circle is (x + 2)² + (y – 3)² = 5
x² + 4x + 4 + y² – 6y + 9 – 5 = 0
i.e., x² + y² + 4x – 6y + 8 = 0

Question 21.
Find the pole of 3x + 4y – 45 = 0 with respect to x² + y² – 6x – 8y + 5 = 0. [A.P. Mar. 16]
Solution:
Equation of polar is
xx₁ + yy₁ – 3(x + x₁) – 4(y + y₁) + 5 = 0
x(x₁ – 3) + y(y₁ – 4) – 3x₁ – 4y₁ + 5 = 0 …………………. (i)
Polar equation is same 3x + 4y – 45 = 0 ……………….. (ii)
Comparing (i) and (ii) we get

Question 22.
i) Show that the circles x² + y² – 6x – 2y + 1 = 0 ; x² + y² + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact. [A.P. Mar. 16; Mar. 11]
Solution:
Equations of the circles are
S₁ ≡ x² + y² – 6x – 2y + 1 = 0
S₂ ≡x² + y² + 2x – 8y + 13 = 0
Centres are A (3, 1), B(-1, 4)
r₁ = \(\sqrt{9+1+1}\) = 3, r₁ = \(\sqrt{1+16-13}\) = 2
AB = \(\sqrt{(3+1)^{2}+(1-4)^{2}}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5
AB = 5 = 3 + 2 = r₁ + r₁
∴ The circles touch each other externally.
The point of contact P divides AB internally in the ratio r₁ : r₂ = 3 : 2
Co- ordinates of P are
\(\left(\frac{3(-1)+2.3}{5}, \frac{3.4+2.1}{5}\right)\) i.e., P\(\left(\frac{3}{5}, \frac{14}{5}\right)\)
Equation of the common tangent is S₁ – S₂ = 0
-8x + 6y – 12 = 0 (or) 4x – 3y + 6 = 0

ii) Show that x² + y² – 6x – 9y + 13 = 0, x² + y² – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Equations of the circles are
S₁ ≡ x² + y² – 6x – 9y + 13 = 0
S₂ ≡ x² + y² – 2x – 16y = 0
centres are A(3, \(\frac{9}{2}\)), B(1, 8)
r₁ = \(\sqrt{9+\frac{81}{4}-13}\) = \(\frac{\sqrt{65}}{2}\), r₂ = \(\sqrt{1+64}\)
= \(\sqrt{65}\)
AB = \(\sqrt{(3-1)^{2}+\left(\frac{9}{2}-8\right)^{2}}\) = \(\sqrt{4+\frac{49}{4}}\)
= \(\frac{\sqrt{65}}{2}\)
AB = |r₁ – r₂|
∴ The circles touch each other internally. The point of contact ‘P’ divides AB
externally in the ratio r₁ : r₂ = \(\frac{\sqrt{65}}{2}\) : \(\sqrt{65}\)
= 1 : 2 Co-ordinates of P are
\(\left(\frac{1(1)-2(3)}{1-2}, \frac{1(8)-2\left(\frac{9}{2}\right)}{1-2}=\left(\frac{-5}{-1}, \frac{-1}{-1}\right)\right.\) = (5, 1)
p = (5, 1)
∴ Equation of the common tangent is
S₁ – S₂ = 0
-4x + 7y + 13 = 0
4x – 7y – 13 = 0

Question 23.
Find the direct common tangents of the circles. [T.S. Mar. 15]
x² + y² + 22x – 4y – 100 = 0 and x² + y² – 22x + 4y + 100 = 0.
Solution:
C₁ = (-11, 2)
C₂ = (11, -2)
r₁ = \(\sqrt{121+4+100}\) = 15
r₂ = \(\sqrt{121+4-100}\) = 5
Let y = mx + c be tangent
mx – y + c = 0
⊥ from (-11, 2) to tangent = 15
⊥ from (11 ,-2) to tangent = 5

Squaring and cross multiplying
25 (1 + m²) = (11 m + 2 – 22m – 4)²
96m² + 44m – 21 = 0
⇒ 96m² + 72m – 28m – 21 = 0
m = \(\frac{7}{24}\), \(\frac{-3}{4}\)
c = \(\frac{25}{2}\)
y = – \(\frac{3}{4}\)x + \(\frac{25}{2}\)
4y + 3x = 50
c = -22m – 4
= -22(\(\frac{7}{24}\)) – 4
= \(\frac{-77-48}{12}\) =\(\frac{-125}{12}\)
y = \(\frac{7}{24}\) x – \(\frac{125}{12}\)
⇒ 24y = 7x – 250
⇒ 7x – 24y – 250 = 0

Question 24.
Find the transverse common tangents of the circles x² + y² – 4x – 10y + 28 = 0 and x² + y² + 4x-6y + 4 = 0 [A.P. Mar. 15; Mar. 14]
Solution:
C₁ =(2, 5), C₂ = (-2, 3)
r₁ = \(\sqrt{4+25-28}\) = 1,
r₂ = \(\sqrt{4+9-4}\) = 3
r₁ + r₂= 4
C₁C₂ = \(\sqrt{(2+2)^{2}+(5-3)^{2}}\)
= \(\sqrt{16+4}\) = \(\sqrt{20}\)
‘C’ divides C₁C₂ in the ratio 5 : 3

Equation of the pair transverse of the common tangents is
S₁² = SS₁₁
(x . 1 + \(\frac{9}{2}\)y – 2(x + 1) – 5(y + \(\frac{9}{2}\)) + 28)² = [1 + \(\frac{81}{4}\) – 4 – 10 × \(\frac{9}{2}\) + 28]
= – (x² + y² – 4x – 10y + 28)
⇒ (-x – \(\frac{1}{2}\)y + \(\frac{7}{2}\))²
= \(\frac{1}{4}\) (x² + y² – 4x – 10y + 28)
(-2x – y + 7)² = (x² + y² – 4x – 10y + 28)
4x² + y² + 4xy – 28x – 14y + 49 = x² + y² – 4x – 10y + 28
3x² + 4xy – 24x – 4y + 21 = 0
(3x + 4y – 21); (x – 1) = 0
3x + 4y – 21 = 0; x – 1 = 0

Question 25.
Show that the circles x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0 touch each other. Also find the point of contact and common tangent at this point of contact. [Mar. 13]
Solution:
Equations of the circles are
x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0
Centres are C₁(2, 3), C₂ = (-3, -9)
r₁ = \(\sqrt{4+9+12}\) = 5
r₂ = \(\sqrt{9+81-26}\) = 8
C₁C₂ = \(\sqrt{(2+3)^{2}+(3+9)^{2}}\)
= \(\sqrt{25+144}\) = 13 = r₁ + r₂
∴ Circle touch externally .
Equation of common tangent is S₁ – S₂ = 0
-10x -,24y – .38 = 0
5x + 12y + 19 = 0

Question 26.
Find the equation of circle with centre (1, 4) and radius ‘5’.
Solution:
Here (h, k) = (1, 4) and r = 5.
∴ By the equation of the circle with centre at C (h, k) and radius r is
(x – h)² + (y – k)² = r²
(x – 1)² + (y – 4)² = 5²
i.e., x² + y²– 2x – 8y – 8 = 0

Question 27.
Find the centre and radius of the circle x² + y² + 2x – 4y – 4 = 0.
Solution:
2g = 2, 2f = -4, c = -4
g = 1, f = -2, c = -4
Centre (-g, -f) = (-1, 2)
radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{1+4-(-4)}\) = 3

Question 28.
Find the centre and radius of the circle 3x² + 3y² – 6x + 4y – 4 = 0.
Solution:
Given equation is
3x² + 3y² – 6x + 4y – 4 = 0
Dividing with 3, we have
x² + y² – 2x + \(\frac{4}{3}\) y – \(\frac{4}{3}\) = 0
2g = -2, 2f = \(\frac{4}{3}\), c = –\(\frac{4}{3}\)
g = -1, f = \(\frac{2}{3}\), c = –\(\frac{4}{3}\)
Centre (-g, -f) = (1, \(\frac{-2}{3}\))
radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{1+\frac{4}{9}+\frac{4}{3}}\)
= \(\sqrt{\frac{9+4+12}{9}}\) = \(\sqrt{\frac{25}{9}}\) = \(\frac{5}{3}\)

Question 29.
Find the equation of the circle whose centre is (-1, 2) and which passes through (5, 6).
Solution:
Let C(-1, 2) be the centre of the circle

Since P(5,6) is a point on the circle CP = r
CP² = r² ⇒ r² = (-1 – 5)² + (2 – 6)²
= 36 + 16 = 52
Equation of the circle is (x + 1)² + (y – 2)²
= 52
x² + 2x + 1 + y² – 4y + 4 – 52 = 0
x² + y² + 2x – 4y – 47 = 0

Question 30.
Find the equation of the circle passing through (2, 3) and concentric with the circle x² + y² + 8x + 12y + 15 = 0.
Solution:
The required circle is concentric with the circle x² + y² + 8x + 12y + 15 = 0
∴ The equation of the required circle can be taken as
x² + y² + 8x + 12y + c = 0

This circle passes through P(2, 3)
∴ 4 + 9 + 16 + 36 + c = 0
c = – 65
Equation of the required circle is x² + y² + 8x + 12y – 65 = 0

Question 31.
From the point A(0, 3) on the circle x² + 4x + (y – 3)² = 0 a chord AB is drawn and extended to a point M such that AM = 2 AB. Find the equation of the locus of M.
Solution:
Let M = (x’, y’)
Given that AM = 2AB

AB + BM = AB + AB
⇒ BM = AB
B is the mid point of AM
Co- ordinates of B are \(\left(\frac{x^{\prime}}{2}-\frac{y^{\prime}+3}{2}\right)\)
B is a point on the circle
(\(\frac{x^{\prime}}{2}\))² + 4(\(\frac{x^{\prime}}{2}\)) + (\(\frac{y^{\prime}+3}{2}\) – 3)² = 0
\(\frac{x^{\prime 2}}{4}\) + 2x’ + \(\frac{y^{\prime 2}-6 y^{\prime}+9}{4}\) = 0
x’² + 8x’ + y’² – 6y’ + 9 = 0
Lotus of M(x’, y’) is x² + y² + 8x – 6y + 9 = 0, which is a circle.

Question 32.
If the circle x² + y² + ax + by – 12 = 0 has the centre at (2, 3) then find a, b and the radius of the circle.
Solution:
Equation of the circle is
x² + y² + ax + by – 12 = 0
Centre = (\(-\frac{a}{2}\), \(-\frac{b}{2}\)) = (2, 3)
\(-\frac{a}{2}\) = 2, \(-\frac{b}{2}\) = 3
a = – 4, b – -6
g = -2, f = -3, c = -12
radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{4+9+12}\) = 5

Question 33.
If the circle x² + y² – 4x + 6y + a = 0 has radius 4 then find a. [A.P. Mar. 16]
Solution:
Equation of the circle is x² + y² – 4x + 6y + a = 0
2g = – 4, 2f = 6, c = a
g =-2, f = 3, c = a
radius = 4 ⇒ \(\sqrt{g^{2}+f^{2}-c}\) = 4
\(\sqrt{4+9-a}\) = 4
13 – a = 16
a = 13 – 16 = -3

Question 34.
Find the equation of the circle passing through (4, 1), (6, 5) and having the centre on the line 4x + y – 16 = 0.
Solution:
Let the equation of the required circle be x² + y² + 2gx + 2fy + c = 0 .
This circle passes through A(4, 1)
16 + 1+ 8g + 2f + c = 0
8g + 2f + c = -17 ………………. (1)
The circle passes through B(6, 5)
36 + 25 + 12g + 10f + c = 0
12g + 10f + c = -61 ……………… (2)
The centre (-g, -f) lies on 4x + y – 16 = 0
– 4g – f – 16 = 0
4g + f + 16 = 0 ……………….. (3)
(2) – (1) gives 4g + 8f = -44 ……………….. (4)
4g + f = -16 …………….. (3)
7f = -28
f = \(\frac{-28}{7}\) = -4
From(3) 4g – 4 = -16
4g = -12 ⇒ g = -3
From(1) 8(-3) + 2(-4) + c = -17
c = -17 + 24 + 8 = 15
Equation of the required circle is
x² + y² – 6x – 8y + 15 = 0

Question 35.
Suppose a point (x₁, y₁) satisfies x² + y²+ 2gx + 2fy + c = 0 then show that it represents a circle whenever g, f and c are real.
Solution:
Comparing with the general equation of second degree co-efficient of x² = coefficient of y² and coefficient of xy = 0
The given equation represents a circle if g² + f² – c ≥ 0
(x₁, y₁) is a point on the given equation
x² + y² + 2gx + 2fy + c = 0, we have
x₁² + y₁² + 2gx₁ + 2fy₁ + c = 0
g² + f² – c = g² + f² + x₁² + y₁² + 2gx₁ + 2fy₁ = 0
= (x₁ + g)² + (y₁ + f)² ≥ 0
g, f and c are real
∴ The given equation represents a circle.

Question 36.
Find the equation of the circle whose extremities of a diameter are (1, 2) and (4, 5).
Solution:
Here (x₁, y₁) = (1, 2) and (x₂, y₂) = (4, 5)
Equation of the required circle is
(x – 1) (x – 4) + (y – 2) ( y – 5) = 0
x² – 5x + 4 + y² – 7y + 10 = 0
x² + y² – 5x – 7y + 14 = 0

Question 37.
Find the other end of the diameter of the circle x² + y² – 8x – 8y + 27 = 0 if one end of it is (2, 3).
Solution:
A = (2, 3) and AB is the diameter of the circle x² + y² – 8x – 8y + 27 = 0

Centre of the circle is C = (4, 4)
Suppose B(x, y) is the other end
C = mid point of AB = \(\left(\frac{2+x}{2}, \frac{3+y}{2}\right)\) = (4, 4)
\(\frac{2+x}{2}\) = 4
2 + x = 8
x = 6
\(\frac{3+y}{2}\) = 4
3 + y = 8
y = 5
The other end of the diameter is B(6, 5)

Question 38.
Find the equation of the circum – circle of the traingle formed by the line ax + by + c = 0 (abc ≠ 0) and the co-ordinate axes.
Solution:
Let the line ax + by + c = 0 cut X, Y axes at A and B respectively co-ordinates of O are (0, 0) A are

Suppose the equation of the required circle is x² + y² + 2gx + 2fy + c = 0
This circle passes through 0(0, 0)
∴ c = 0
This circle passes through A(-\(\frac{c}{a}\), 0)
\(\frac{c^{2}}{a^{2}}\) + 0 – 2\(\frac{\mathrm{gc}}{\mathrm{a}}\) = 0
2g . \(\frac{c}{a}\) = \(\frac{c^{2}}{a^{2}}\) ⇒ 2g = \(\frac{c}{a}\) ⇒ g = \(\frac{c}{2a}\)
The circle passes through B (0, –\(\frac{c}{b}\))
0 + \(\frac{c^{2}}{b^{2}}\) + 0 – 2g \(\frac{c}{b}\) = 0
2f\(\frac{c}{b}\) = \(\frac{c^{2}}{b^{2}}\) ⇒ 2g = \(\frac{c}{b}\) ⇒ f = \(\frac{c}{2b}\)
Equation of the circle, through O, A, B is
x² + y² + \(\frac{c}{a}\) x + \(\frac{c}{b}\) y = 0
ab(x² + y²) + (bx + ay) = 0
This is the equation of the circum circle of ∆OAB

Question 39.
Find the equation of the circle which passes through the vertices of the triangle formed by L₁ = x + y + 1 = 0; L₂ = 3x + y- 5 = 0 and L₃ = 2x + y – 5 = 0.
Solution:
Suppose L₁, L₂,: L₂, L₃ and L₃, L₁ intersect in A, B and C respectively.
Consider a curve whose equation is
k (x + y + 1) (3x + y – 5) + l(3x + y – 5) (2x + y – 5) + m(2x + y – 5) (x + y + 1) = 0 ………………. (1)
This equation represents a circle
i) Co-efficient of x² = Co – efficient of y²
3k + 6l + 2m = k + l + m
2k + 5l + m = 0 ……………….. (2)
ii) Co-efficient of xy = 0
4k + 5l + 3m = 0 ……………….. (3).
Applying cross multiplication rule for (2) and (3) we get

Substituting in (1), equation of the required circle is
5(x + y + 1) (3x + y – 5) – 1 (3x + y – 5)
(2x + y – 5) – 5(2x + y – 5) (x + y + 1) = 0
i.e., x² + y² – 30x – 10y + 25 = 0

Question 40.
Find the centre of the circle passing through the points (0, 0), (2, 0)and (0, 2).
Solution:
Here (x₁, y₁) = (0, 0); (x₂, y₁) = (2, 0);
(x₃, y₃) = (0, 2)
c₁ = -(x₁² + y₁²) = 0
c₂ = – (x₂² + y₂²) = -(2² + 0²) = -4
c₃ = -(x₃² + y₃²) = -(0² + 2²) – 4
The centre of the circle passing through three non-collinear points P(x₁, y₁), Q(x₂, y₂) and R(x₃, y₃)

Thus the centre of the required circle is (1, 1)

Question 41.
Obtain the parametric equations of the circle x² + y² = 1.
Solution:
Equation of the circle is x² + y² = 1
Centre is (0, 0) radius = r = T

The circle having radius r is x = r cos θ,
y = sin θ where 0 < θ < 2π
The parametric equation of the circle
x² + y² = 1 and
x = 1 . cos θ = cos θ
y = 1 . sin θ = sin θ, θ < θ < 2π
Note: Every point on the circle can be expressed as (cos θ, sin θ)

Question 42.
Obtain the parametric equation of the circle represented by
x² + y² + 6x + 8y – 96 = 0.
Solution:
Centre (h, k) of the circle is (-3, -4)
radius = r = \(\sqrt{9+16+96}\) = \(\sqrt{121}\) = 11
Parametric equations are
x = h + r cos θ = -3 + 11 cos θ
y = k + r sin θ = -4 + 11 sin θ
where 0 < θ < 2π

Question 43.
Locate the position of the point (2, 4) with respect to the circle. x² + y² – 4x – 6y + 11 = 0.
Solution:
Here (x₁, y₁) = (2, 4) and
S ≡ x² + y² – 4x – 6y + 11
S₁₁ = 4 + 16 – 8 – 24 + 11
= 31 – 32 = – 1 < 0
∴ The point (2, 4) lies inside the circle S = 0

Question 44.
Find the length of the tangent from (1, 3) to the circle x² + y² – 2x + 4y – 11 = 0.
Solution:
Here (x₁, y₁) = (1, 3) and
S = x² + y² – 2x + 4y – 11 = 0
P(x₁, y₁) to S = 0 is \(\sqrt{S_{11}}\)
Length of the tangent = \(\sqrt{S_{11}}\)
= \(\sqrt{1+9-2+12-11}\) = \(\sqrt{9}\) = 3

Question 45.
If a point P is moving such that the length of tangents drawn from P to
x² + y² – 2x + 4y – 20 = 0 ……………… (1)
and x² + y² – 2x – 8y + 1 = 0 ……………….. (2)
are in the ratio 2 : 1.
Then show that the equation of the locus of P is x² + y² – 2x – 12y + 8 = 0.
Solution:
Let P(x₁, y₁) be any point on the locus and \(\overline{\mathrm{PT}_{1}}\), \(\overline{\mathrm{PT}_{2}}\) be the lengths of tangents from P to the circles (1) and (2) respectively.
x² + y² – 2x + 4y – 20 = 0 and
x² + y² – 2x – 8y + 1 = 0
\(\frac{\overline{\mathrm{PT}_{1}}}{\overline{\mathrm{PT}_{2}}}=\frac{2}{1}\)
i.e., \(\sqrt{x_{1}^{2}+y_{1}^{2}-2 x_{1}+4 y_{1}-20}\)
= \(2 \sqrt{x_{1}^{2}+y_{1}^{2}-2 x_{1}-8 y_{1}+1}\)
3 (x₁² + y₁²) – 6x₁ – 36y₁ + 24 = 0
Locus of P (x₁, y₁) is
x² + y² – 2x – 12y + 8 = 0

Question 46.
If S ≡ x² + y² + 2gx + 2fy + c = 0. represents a circle then show that the straight line lx + my + n = 0
i) touches the circle S = 0 if
(g² + f² – c) = \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

ii) meets the circle S = 0 in two points if
g² + f² – c > \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

iii) will not meet the circle if
g² + f² – c < \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)
Solution:
Let ‘c’ be the centre and ‘r’ be the radius of the circle S = 0
Then C = (-g, -f) and r = \(\sqrt{g^{2}+f^{2}-c}\)
i) The given straight line touches the circle
if r = \(\frac{|l(-\mathrm{g})+\mathrm{m}(-\mathrm{f})-\mathrm{n}|}{\sqrt{l^{2}+\mathrm{m}^{2}}}\)
\(\sqrt{g^{2}+f^{2}-c}\) = \(\frac{|-(l g+m f-n)|}{\sqrt{l^{2}+m^{2}}}\)
squaring on both sides, we get
g² + f² – c = \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

ii) The given line lx + my + n = 0 meets the circle s = 0 in two points if
g² + f² – c > \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

iii) The given line lx + my + n = 0 will not meet the circle s = 0 if
g² + f² – c < \(\frac{(g l+m f-n)^{2}}{\left(l^{2}+m^{2}\right)}\)

Question 47.
Find the length of the chord intercepted by the circle x² + y² + 8x – 4y – 16 = 0 on the line 3x – y + 4 = 0.
Solution:
The centre of the given circle c = (-4, 2) and radius r = \(\sqrt{16+4+16}\) = 6. Let the perpendicular distance from the centre to the line 3x-y + 4 = 0 be ‘d’ then
d = \(\frac{|3(-4)-2+4|}{\sqrt{3^{2}+(-1)^{2}}}\) = \(\frac{10}{\sqrt{10}}\) = \(\sqrt{10}\)
Length of the chord + \(\sqrt{r^{2}-d^{2}}\)
= 2\(\sqrt{6^{2}-(\sqrt{10})^{2}}\) = 2\(\sqrt{26}\)

Question 48.
Find the equation of tangents to x² + y² – 4x + 6y – 12 = 0 which are parallel to x + 2y- 8 = 0.
Solution:
Here g = -2, f = 3, r = \(\sqrt{4+9+12}\) = 5
and the slope of the required tangent is \(\frac{-1}{2}\)
The equations of tangents are
y + 3 = \(\frac{-1}{2}\) (x – 2) ± 5 \(\sqrt{1+\frac{1}{4}}\)
2(y + 3) = – x + 2 ± 5\(\sqrt{5}\)
x + 2y + (4 ± 5\(\sqrt{5}\)) = 0

Question 49.
Show that the circle S ≡ x² + y² + 2gx + 2fy + c = 0 touches the
i) X – axis if g² = c
ii) Y – axis if f² = c.
Solution:
i) We know that the intercept made by S = 0 on X – axis is 2\(\sqrt{g^{2}-c}\)
If the circle touches the X – axis then
22\(\sqrt{g^{2}-c}\) 0 ⇒ g² = c

Question 50.
Find the equation of the tangent to x² + y² – 6x + 4y – 12 = 0 at (- 1, 1).
Solution:
Here (x₁, y₁) = (-1, 1) and
S ≡ x² + y² – 6x + 4y – 12 = 0
∴ The equation of the tangent is
x(-1) + y (1) – 3(x – 1) + 2(y + 1)- 12 = 0
The equation of the tangent at the point P(₁, y₁) to the circle
S ≡ x² + y² + 2gx + 2fy + c = 0 is S₁ = 0
⇒ – x + y – 3x + 3 + 2y + 2 – 12 = 0
⇒ – 4x + 3y – 7 = 0
(or) 4x – 3y + 7 = 0

Question 51.
Find the equation of the tangent to x² + y² – 2x + 4y = 0 at (3, -1). Also find the equation of tangent parallel to it.
Solution:
Here (x₁, y₁) = (3, -1) and
S ≡ x² + y² – 2x + 4y = 0
The equation of the tangent at (3, -1) is
x (3) + y (-1) – (x + 3) + 2(y – 1) = 0
3x – y – x – 3 + 2y – 2 = 0
2x + y – 5 = 0
Slope of the tangent is m = -2, for the circle
g = -1, f = 2, c = 0
r = \(\sqrt{1+4-0}\) = \(\sqrt{5}\)
Equations of the tangents are
y = mx ± r \(\sqrt{1+m^{2}}\) is a tangent to the circle x² + y² = r² and the slope of the tangent is m.
y + 2 = -2(x – 1) ± \(\sqrt{5} \sqrt{1+4}\)
y + 2 = -2x + 2 ± 5
2x 4- y = ± 5
The tangents are
2x + y + 5 = 0 and 2x + y – 5 = 0
The tangent parallel to the given tangent is 2x + y ± 5 = 0

Question 52.
If 4x – 3y + 7 = 0 is a tangent to the circle represented by x² + y² – 6x + 4y – 12 = 0, then find its point of contact.
Solution:
Let (x₁, y₁) be the point of contact
Equation of the tangent is
(x₁ + g) x + (y₁ + f) y + (gx₁ + fy₁ + c) = 0
We have \(\frac{x_{1}-3}{4}\) = \(\frac{y_{1}+2}{-3}\)
= \(\left(\frac{-3 x_{1}+2 y_{1}-12}{7}\right)\) ……………… (1)
From first and second equalities of (1), we get
3x₁ + 4y₁ = 1 ………………. (2)
Now by taking first and third equalities of (1), we get
19x₁ – 8y₁ = -27 ………………. (3)
Solving (2) and (3) we obtain
x₁ = -1;, y₁ = 1
Hence the point of contact is (-1, 1).

Question 53.
Find the equations of circles which touch 2x – 3y + 1 = 0 at (1, 1) and having radius \(\sqrt{13}\).
Solution:
The centres of the required circle lies on a line perpendicular to 2x – 3y + 1 =0 and passing through (1, 1)

The equation of the line of centre can be taken as
3x + 2y + k = 0
This line passes through (1, 1)
3 + 2 + k = 0 ⇒ k = -5
Equation of AB is 3x + 2y – 5 = 0
The centres A and B are situated on
3x + 2y – 5 = 0 at a distance \(\sqrt{13}\) from (1, 1).
The centres are given by
(x₁ ± r cos θ, y₁ ± r sin θ)
\(\left(1+\sqrt{13}\left(-\frac{2}{\sqrt{13}}\right) 1+\sqrt{13} \cdot \frac{3}{\sqrt{13}}\right)\) and
\(\left(1-\sqrt{13} \frac{(-2)}{\sqrt{13}}, 1-\sqrt{13} \cdot \frac{3}{\sqrt{13}}\right)\)
i.e., (1 -2, 1 +3) and (1 + 2, 1 – 3)
(-1, 4) and (3, -2)
Centre (3, -2), r = \(\sqrt{13}\)
Equation of the required circles are
(x + 1 )² + (y – 4)² = 13 and
(x – 3)² + (y + 2)² = 13
i.e., x² + y² + 2x – 8y + 4 = 0
and x² + y² – 6x + 4y = 0

Question 54.
Show that the line 5x + 12y – 4 = 0 touches the circle x² + y² – 6x + 4y + 12 = 0.
Solution:
Equation of the circle is
x² + y² – 6x + 4y + 12 = 0
Centre (3, -2), r = \(\sqrt{9+4-12}\) = 1 ……………….. (1)
The given line touches the circle if the perpendicular distance from the centre on the given line is equal to radius of the circle, d = perpendicular distance from (3, -2)
= \(\frac{|5(3)+12(-2)-4|}{\sqrt{25+144}}\)
= \(\frac{13}{13}\) = 1 = radius of the circle ………………… (2)
∴ The given line 5x + 12y – 4 = 0 touches the circle.

Question 55.
If the parametric values of two points A and B lying on the circle x² + y² – 6x + 4y – 12 = 0 are 30° and 60° respectively, then find the equation of the chord joining A and B.
Solution:
Here g = -3, f = -2; r = \(\sqrt{9+4-12}\) = 5
∴ The equation of the chord joining the points θ₁ = 30°, θ₂ = 60°
Equation of chord joining the point; (-g+ r cos θ₁(-f + r sin θ₁) where r is the radius of the circle; θ₂ and (-g + r cos θ₂, -f + r sin θ₂) is (x + g) cos (\(\frac{\theta_{1}+\theta_{2}}{2}\)) + (y + f)
sin (\(\frac{\theta_{1}+\theta_{2}}{2}\)) = r cos (\(\frac{\theta_{1}+\theta_{2}}{2}\))
(\(\frac{\theta_{1}+\theta_{2}}{2}\)) = r cos \(\frac{\theta_{1}+\theta_{2}}{2}\)
(x – 3) cos [latex]\frac{60^{\circ}+30^{\circ}}{2}[/latex]
(y + 2) sin [latex]\frac{60^{\circ}+30^{\circ}}{2}[/latex]
= 5 cos [latex]\frac{60^{\circ}-30^{\circ}}{2}[/latex]
i.e., (x – 3) cos 45 ° + (y + 2) sin 45°
= 5 cos 15°
= \(\frac{(x-3)+(y+2)}{\sqrt{2}}=5\left[\frac{(\sqrt{3}+1)}{2 \sqrt{2}}\right]\)
i.e., 2x + 2y – (7 + 5\(\sqrt{3}\)) = 0.

Question 56.
Find the equation of the tangent at the point 30° (parametric value of θ) of the circle is x² + y² + 4x + 6y – 39 = 0.
Solution:
Equation of the circle is
x² + y² + 4x + 6y – 39 = 0
g = 2,f = 3, r = \(\sqrt{4+9+39}\) = \(\sqrt{52}\) = 2\(\sqrt{3}\)
θ = 30°
Equation of the tangent is
(x + g) cos 30° + (y + f) sin 30° = r
(x + 2) \(\frac{\sqrt{3}}{2}\) +(y + 3) \(\frac{1}{2}\) = 2713
\(\sqrt{3}\)x + 2\(\sqrt{3}\) + y + 3 = 4\(\sqrt{13}\)
\(\sqrt{3}\) x + y + (3 + 2\(\sqrt{3}\) – 4\(\sqrt{13}\)) = 0

Question 57.
Find the area of the triangle formed by the tangent at P(x₁, y₁) to the circle x² + y² = a² with the co-ordinate axes where x₁y₁ ≠ 0.
Solution:
Equation of the circle is x² + y² = a²
Equation of the tangent at P(x₁, y₁) is xx₁ + yy₁ = a² ……………… (1)

= \(\frac{a^{4}}{2\left|x_{1} y_{1}\right|}\)

Question 58.
Find the equation of the normal to the circle x² + y² – 4x – 6y + 11 = 0 at (3, 2). Also find the other point where the normal meets the circle.
Solution:
Let A(3, 2), C be the centre of given circle and the normal at A meet the circle at B(a, b). From the hypothesis, we have
2g = -4 i.e., g = -2
2f = -6 i.e., f = -3
x₁ = 3 and y₁ = 2
The equation of the normal at P(x₁, y₁) of the circle
S ≡ x² + y² + 2gx + 2fy + c = Q is
(x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0
The equation of normal at A(3, 2) is
(x – 3) (2 – 3) – (y – 2) (3 – 2) = 0
i.e., x + y – 5 = 0
The centre of the circle is the mid point of A and B. (Points of intersection of normal and circle).
\(\frac{a+3}{2}\) = 2 ⇒ a = 1
and \(\frac{b+2}{2}\) = 3 ⇒ b = 4
Hence the normal at (3, 2) meets the circle at (1, 4).

Question 59.
Find the area of the triangle formed by the normal at (3, -4) to the circle x² + y² – 22x – 4y + 25 = 0 with the co-ordinate axes.
Solution:
Here 2g = -22, 2f = – 4, g = -11, f = -2
x₁ = 3, y₁ = – 4
Equation of the normal at (3, -4) is
(x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0
(x – 3) (-4 – 2) – (y + 4) (3 – 11) = 0
3x + 4y – 25 = 0 ……………….. (1)
This line meets X-axis at A(\(\frac{25}{3}\), 0) and Y – axis at B(0, \(\frac{25}{4}\)) , ∆OAB = \(\frac{1}{2}\) |OA . OB|
= \(\frac{1}{2}\) |\(\frac{25}{3} \times-\left[\frac{25}{4}\right]\)| = \(\frac{625}{24}\)

Question 60.
Show that the line lx + my + n =0 is a normal to the circle S = 0 if and only if gl + mf = n.
Solution:
The straight line lx + my + n = 0 is normal to the circle
S = x² + y² + 2gx + 2fy + c = 0
⇒ If the centre (-g, -f) lies on
lx + my + n = 0
l (- g) + m(- f) + n = 0
gl + fm = n

Question 61.
Find the condition that the tangents drawn from the exterior point (g, f) to S ≡ x² + y² + 2gx + 2fy + c = 0 are perpendicular to each other.
Solution:
If the angle between the tangents drawn from P(x₁, y₁) to S = 0 is θ, then
tan (\(\frac{\theta}{2}\)) = \(\frac{\mathrm{r}}{\sqrt{\mathrm{S}_{11}}}\)
Equation of the circle is
x² + y² + 2gx + 2fy + c = 0
r = \(\sqrt{g^{2}+f^{2}-c}\)
S₁₁ = g² + f² + 2g² + 2f² + c
= 3g² + 3f² + c
θ = 90° ⇒ tan \(\frac{\theta}{2}\) = tan 45 = \(\frac{\sqrt{g^{2}+f^{2}-c}}{\sqrt{3 g^{2}+3 f^{2}+c}}\)
1 = \(\frac{g^{2}+f^{2}-c}{3 g^{2}+3 f^{2}+c}\)
⇒ 3g² + 3f² + c = g² + f² – c
⇒ 2g² + 2f² + 2c = 0 ⇒ g² + f² + c = 0
This is the condition for the tangent drawn from (g, f) to the circle S = 0 are perpendicular.
Note : Here c < 0

Question 62.
If θ₁, θ₂ are the angles of inclination of tangents through a point P to the circle x² + y² = a² then find the locus of P when cot θ₁ + cot θ₂ = k.
Solution:
Equation of the circle is x² + y² = a²
If m is the slope of the tangent, then the equation of tangent passing through P(x₁, y₁) can be taken as
y₁ = mx₁ ± a\(\sqrt{1+m^{2}}\)
(y₁ – mx₁)² = a² (1 + m²)
m²x₁² + y₁² – 2mx₁y₁ – a² – a²m² = 0
m² (x₁² – a²) – 2mx₁y₁ + (y₁² – a²) = 0
Suppose m₁ and m₂ are the roots of this equation
m₁ + m₂ = tan θ₁ + tan θ₂ = \(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}\)
m₁m₂ – tan θ₁ . tan θ₁ = \(\frac{y_{1}^{2}-a^{2}}{x_{1}^{2}-a^{2}}\)
Given that cot θ₁ + cot θ₂ = k
⇒ \(\frac{1}{\tan \theta_{1}}+\frac{1}{\tan \theta_{2}}\) = k
⇒ \(\frac{\tan \theta_{1}+\tan \theta_{2}}{\tan \theta_{1} \cdot \tan \theta_{2}}\) = k
⇒ \(\frac{2 x_{1} y_{1}}{y_{1}^{2}-a^{2}}\) = k
2x₁y₁ = k (y₁² – a²)
Locus of P(x₁, y₁) is 2xy = k(y² – a²)
Also, conversely if P(x₁, y₁) satisfies the condition 2xy = k(y² – a²) then it can be show that cot θ₁ + cot θ₂ = k
Thus the locus of P is 2xy = k(y² – a²)

Question 63.
Find the chord of contact of (2, 5) with respect to the circle x² + y² – 5x + 4y – 2 = 0.
Solution:
Here (x₁, y₁) = (2, 5). By
S ≡ x² + y² + 2gx + 2fy + c = 0 then the equation of the chord of contact of P with respect to S = 0 is S₁ =0, the required chord of contact is
xx₁ + yy₁ – \(\frac{5}{2}\) (x + x₁) + 2(y + y₁) – 2 = 0
Substituting x₁ and y₁ values, we get
x(2) + y(5) – \(\frac{5}{2}\) (x + 2) + 2(y + 5) – 2 = 0
i.e., x – 14y + 6 = 0.

Question 64.
If the chord of contact of a point P with respect to the circle x² + y² = a² cut the circle at A and B such that, AÔB = 90° then show that P lies on the circle x² + y² = 2a².
Solution:
Given circle x² + y² = a² …………… (1)
Let P(x₁, y₁) be a point and let the chord of contact of it cut the circle in A and B such that AÔB = 90°. The equation of the chord of contact of P(x₁, y₁) with respect to (1) is
xx₁ + yy₁ – a² = 0 ……………………. (2)
The equation of the pair of the lines \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) is.given by x² + y² – a²
\(\left(\frac{x x_{1}+y y_{1}}{a^{2}}\right)^{2}\) = 0
or a² (x² + y²)- (xx₁ + yy₁)² = 0
or x² (a² – x₁²) – 2x₁y₁xy + y² (a² – y₁²) = 0 – (3)
Since AÔB = 90°, we have the coefficient of x² in (3) + coefficient of y² in (3) = 0
∴ a² – x₁² + a² – y₁² = 0
i.e., x₁² + y₁² = 2a²
Hence the point P(x₁, y₁) lies on the circle x² + y² = 2a².

Question 65.
Find the equation of the polar of (2, 3) with respect to the circle x² + y² + 6x + 8y – 96 = 0.
Solution:
Here (x₁, y₁) = (2, 3) ⇒ x₁ = 2, y₁ = 3
Equation of the circle is
x² + y² + 6x + 8y – 96 = 0
Equation of the polar is S₁ = 0
Polar of (2, 3) is x . 2 + y . 3 + 3(x + 2) + 4(y + 3) – 96 = 0
2x + 3y + 3x + 6 + 4y + 12 – 96 = 0
5x + 7y – 78 = 0

Question 66.
Find the pole of x + y + 2 = 0 with respect to the circle x² + y² – 4x + 6y – 12 = 0.
Solution:
Here lx + my + n = 0 is x + y + 2 = 0
l = 1, m = 1, n = 2
Equation of the circle is
S ≡ x² + y² – 4x + 6y – 12 = 0
2g = -4, 2f = 6, c = -12
g = -2, f = 3, c = -12

The pole of the given line x + y + 2 = 0 w.r.to the given circle (-23, -28)

Question 67.
Show that the poles of the tangents to the circle x² + y² = a² with respect to the circle (x + a)² + y² = 2a² lie on y² + 4ax = 0.
Solution:
Equations of the given circles are
x² + y² = a² …………………. (1)
and (x + a)² + y² = 2a² ………………. (2)
Let P(x₁, y₁) be the pole of the tangent to the circle (1) with respect to circle (2).
The polar of P w.r.to circle (2) is
xx₁ + yy₁ + a(x + x₁) – a² = 0
x(x₁ + a) + yy₁ + (ax₁ – a₂) = 0
This is a tangent to circle (1)
∴ a = \(\frac{\left|0 .+0+a x_{1}-a^{2}\right|}{\sqrt{\left(\dot{x}_{1}+a\right)^{2}}+y_{1}^{2}}\)
a = \(\frac{a\left|x_{1}-a\right|}{\sqrt{\left(x_{1}+a\right)^{2}}+y_{1}^{2}}\)
Squaring and cross – multiplying
(x₁ + a)² + y₁² = (x₁ – a)²
(or) y₁² + (x₁ + a)² – (x₁ – a)² = 0
i.e., y₁² + 4ax₁ = 0
The poles of the tangents to circle (1) w.r.to circle (2) lie on the curve y² + 4ax = 0

Question 68.
Show that (4, -2) and (3, -6) are conjugate with respect to the circle x² + y² – 24 = 0.
Solution:
Here (x₁, y₁) = (4, -2) and (x₂, y₂) = (3, -6) and
S ≡ x² + y² – 24 = 0
Two points (x₁, y₁) and (x₂, y₂) are conjugate with respect to the circle S = 0 if S₁₂ = 0;
In this case x₁x₂ + y₁y₂ – 24 = 0
S₁₂ = 4.3 + (-2) (-6) – 24 .
= 12 + 12 – 24 = 0
∴ The given points are conjugate with respect to the given circle.

Question 69.
If (4, k) and (2, 3) are conjugate points with respect to the circle x² + y² = 17 then find k.
Solution:
Here (x₁, y₁) = (4, k) and (x₂, y₂) = (2, 3) and
S ≡ x² + y² – 17 = 0
The given points are conjugate ⇒ S₁₂ = 0
x₁x₂ + y₁y₂ – 17 = 0
4.2 + k. 3 – 17 = 0
3k = 17 – 8 = 9
k = \(\frac{9}{3}\) = 3

Question 70.
Show that the lines 2x + 3y + 11 = 0 and 2x – 2y – 1 = 0 are conjugate with respect to the circle x² + y² + 4x + 6y – 12 = 0.
Solution:
Here l₁ = 2, m₁ = 3, n₁ = 11
l₂ = 2, m₂ = -2, n₂ = -1
and g = 2, f = 3, c = 12
r = \(\sqrt{9+4-12}\) = 1
We know that l₁x + m₁y + n₁ = 0
l₂x + m₂y + n₂ = 0 are conjugate with respect to S = 0
r² (l₁l₂ + m₁m₂) = (l₁g + m₁f – n₁) (l₂g + m₂f – n₂)
r² (l₁l₂ + m₁m₂) – 1(2.2 + 3(- 2)) = 4 – 6 = -2
(l₁g + m₁f – n₁) (l₂g + m₂f – n₂)
= (2.2 + 3.3-11) (2.2-2.3 + 11)
= 2(- 1) = -2 L.H.S. = R.H.S.
Condition for conjugate lines is satisfied
∴ Given lines are conjugate lines.

Question 71.
Show that the area of the triangle formed by the two tangents through P(xv to the circle S ≡ x² + y² + 2gx + 2fy + c = 0 and the chord of contact of P with respect to S = 0 is \(\frac{r\left(S_{11}\right)^{3 / 2}}{S_{11}+r^{2}}\) where r is the radius of the circle.
Solution:
PA and PB are two tangents from P to the circle S = 0
AB is the chord of contact

= \(S_{11} \cdot \frac{r}{\sqrt{S_{11}}} \cdot \frac{S_{11}}{S_{11}+r^{2}}\)
= \(\frac{r\left(S_{11}\right)^{3 / 2}}{S_{11}+r^{2}}\)

Question 72.
Find the mid point of the chord intercepted by
x² + y² – 2x – 10y + 1 = 0 ………………. (1)
on the line x – 2y + 7 = 0. ……………. (2)
Solution:
Let P(x₁, y₁) be the mid point of the chord AB
Equation of the chord is S₁ = S₁₁
xx₁ + yy₁ – 1 (x + x₁) – 5(y + y₁) + 1 = x₁² + y₁² – 2x₁ – 10y₁ + 1 .
x(x₁ – 1) + y(y₁ – 5) – (x₁² + y₁² – x₁ – 5y₁) = 0 ………………. (3)
Equation of the given line is x – 2y – 7 = 0
Comparing (1) and (2)
\(\frac{x_{1}-1}{1}=\frac{y_{1}-5}{-2}=\frac{x_{1}^{2}+y_{1}^{2}-x_{1}-5 y_{1}}{-7}\) = k (say)
x₁ – 1 = k ⇒ x₁ = k + 1
y₁ – 5 = -2(k) ⇒ y₁ = 5 – 2k
x₁² + y₁² – x₁ – 5y₁ = -7k
⇒ (k + 1)² + (5 – 2k)² – (k + 1) – 5(5 – 2k) + 7k = 0
⇒ k² + 2k + 1 + 25 + 4k² – 20k – k – 1 – 25 + 10k + 7k = 0
⇒ 5k² – 2k = 0
⇒ k (5k — 2) = 0 ⇒ k = 0, \(\frac{2}{5}\)
k = 0 ⇒ (x₁, y₁) = (1, 5) and x – 2y + 7
= 1 – 10 + 7 ≠ 0
(1, 5) is not a point on the chord.
k = \(\frac{2}{5}\) (x₁, y₁) = \(\left(\frac{7}{5}, \frac{21}{5}\right)\)

Question 73.
Find the locus of mid-points of the chords of contact of x² + y² = a² from the points lying on the line lx + my + n = 0.
Solution:
Let P = (x₁, y₁) be a point on the locus P is the mid point of the chord of the circle
x² + y² = a²
Equation of the chord is lx + my + n = 0 ……………… (1)
Equation of the circle is x² + y² = a²
Equation is the chord having (x₁, y₁) as mid point of S₁ = S₁₁
xx₁ + yy₁ = x₁² + y₁²
xx₁ + yy₁ – (x₁² + y₁²) = 0 ……………….. (2)
Pole of (2) with respect to the circle in

Locus of P(x₁, y₁) is n(x² + y²) + a²(lx + my) = 0

Question 74.
Show that the four common tangents can be drawn for the circles given by
x² + y² – 14x + 6y + 33 = 0 …………… (1)
and x² + y² + 30x – 2y +1 = 0 ……………… (2)
and find the internal and external centres of similitude. [T.S. Mar. 19]
Solution:
Equations of the circles are
x² + y² – 14x + 6y + 33 = 0
and x² + y² + 30x – 2y + 1 = 0
Centres are A(7, -3), B(-15, 1)
r₁ = \(\sqrt{49+9-33}\) = 5
r₂ = \(\sqrt{225+1-1}\) = 15
AB = \(\sqrt{(7+15)^{2}+(-3-1)^{2}}\)
= \(\sqrt{484+16}\) = \(\sqrt{500}\) > r₁ + r₂
∴ Four common tangents can be drawn to the given circle
r₁ : r₂ = 5 : 15 = 1 : 3

Internal centre of similitude S’ divides AB in the ratio 1 : 3 internally
Co-ordinates of S’ are
\(\left(\frac{1 .(-15)+3.7}{1+3}, \frac{1.1+3(-3)}{1+3}\right)\)
= \(\left(\frac{6}{4}, \frac{1-9}{4}\right)\) = \(\left(\frac{3}{2},-2\right)\)
External centre of similitude S divides AB externally in the ratio 1 : 3
Co-ordinates of S are
\(\left(\frac{1(-15)+3(7)}{1-3}, \frac{1.1-3(-3)}{1-3}\right)\)
= \(\left(\frac{-15-21}{-2}, \frac{1+9}{-2}\right)\) = (18, -5)

Question 75.
Prove that the circles x² + y² – 8x – 6y + 21 = 0 and x² + y² – 2y – 15 = 0 have exactly two common tangents. Also find the point of intersection of those tangents.
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be their radii.
Equation of the circles are
x² + y² – 8x – 6y + 21 = 0
and x² + y² – 2y – 15 = 0
C₁(4, 3), C₂(0, 1)
r₁ = \(\sqrt{16+9-21}\) = 2, r₂ = \(\sqrt{1+15}\) = 4
\(\overline{\mathrm{C}_{1} \mathrm{C}_{2}^{2}}\) = (4 – 0)² + (3 – 1)² = 16 + 4 = 20
C₁C₂ = 2\(\sqrt{5}\)
|r₁ – r₂| = |2 – 4| = 2, r₁ + r₂ = 2 + 4 = 6
|r₁ – r₂| < C₁C₂ < r₁ + r₂
Given circles intersect and have exactly two common tangents.
r₁ : r₂ = 2 : 4 = 1 : 2
The tangents intersect in external centre of similitude

Co-ordinates of S are
\(\left(\frac{1.0-2.4}{1-2}, \frac{1.0-2.3}{1-2}\right)\) = \(\left(\frac{-8}{-1}, \frac{-5}{-1}\right)\)
= (8, 5)

Question 76.
Show that the circles x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0 touch each other. Also find the point of contact and common tangent at this point of contact. [Mar. 13]
Solution:
Equations of the circles are
x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0
Centres are C₁(2, 3), C₂ = (-3, -9)
r₁ = \(\sqrt{4+9+12}\) = 5
r₂ = \(\sqrt{9+81-26}\) = 8
C₁C₂ = \(\sqrt{(2+3)^{2}+(3+9)^{2}}\)
= \(\sqrt{25+144}\) = 13 = r₁ + r₂
∴ Circle touch externally
Equation of common tangent is S₁ – S₂ = 0
-10x – 24y – 38 = 0
5x + 12y + 19 = 0

Question 77.
Show that the circles x² + y² – 4x – 6y – 12 = 0 and 5(x² + y²) – 8x – 14y – 32 = 0 touch each other and find their point of contact.
Solution:
Equations of the circles are
x² + y² – 4x – 6y – 12 = 0
and x² + y² – \(\frac{8}{5}\) x – \(\frac{14}{5}\) y – \(\frac{32}{4}\) = 0
Centres are A(2, 3), B(\(\frac{4}{5}\), \(\frac{7}{5}\))

The circles touch internally
P divides AB externally the ratio 5 : 3

Question 78.
Find the equation of the pair of tangents from (10, 4) to the circle x² + y² = 25.
Solution:
(x₁, y₁) = (1o, 4)
Equation of the circle is x² + y² – 25 = 0
Equation of the pair of tangents is S₁² = S . S₁₁
(10x + 4y – 25)² = (100 + 16 – 25)(x² + y² – 25)
100x² + 16y² + 625 + 80xy – 500x – 200y = 91x² + 91y² – 2275
9x² + 80xy – 75y² – 500x – 200y + 2900 = 0

Question 79.
Find the equation to all possible common tangents of the circles x² + y² – 2x – 6y + 6 = 0 and x² + y² = 1.
Solution:
Equations of the circles are
x² + y² – 2x – 6y + 6 = 0
and x² + y² = 1
Centres are A(1, 3), B(0, 0),
r₁ = \(\sqrt{1+9-6}\) = 2
r₂ = 1
External centre of similitude S divides AB externally in the ratio 2 : 1
Co-ordinates of S are
\(\left(\frac{2.0-1.1}{2-1}, \frac{2.0-1.3}{2-1}\right)\) = (-1, -3)
Equation to the pair of direct common tangents are
(x² + y² – 1) (1 + 9 – 1) = (x + 3y + 1)²
This can be expressed as
(y – 1) (4y + 3x – 5) = 0
Equations of direct common tangents are
y – 1 = 0 and 3x + 4y – 5 = 0
Internal centre of S’ divides AB internally in the ratio 2 : 1
Co-ordinates of S’ are
\(\left(\frac{2.0+1.1}{2+1}=\frac{2.0+1.3}{2+1}\right)\left(\frac{1}{3}, 1\right)\)
Equation to the pair of transverse common tangents are
(\(\frac{x}{3}\) +y – 1)²
= (\(\frac{1}{9}\) + 1 – 1) (x² + y² – 1)
This can be expressed as
(x + 1)(4x – 3y – 5) = 0
Equation of transverse common tangents are x + 1 = 0 and 4x – 3y – 5 = 0

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 8 Differential Equations to solve questions creatively.

Intermediate 2nd Year Maths 2B Differential Equations Formulas

→ An equation involving one dependent variable and its derivatives wIth respect to one or more independent variables is called a differential equation.

→ Order of a differential equation is the maximum of the orders of the derivatives.

→ Degree of a differential equation is the degree of the highest order derivative.

→ An equation involving one dependent variable, one or more independent variables, and the differential coefficients (derivatives) of the dependent variable with respect to independent variables is called a differential equation.

→ Order of a Differential Equation:
The order of the highest derivative involved in an ordinary differential equation is called the order of the differential equation.

→ Degree of a Differential Equation:
The degree of the highest derivative involved in an ordinary differential equation, when the equation has been expressed in the form of a polynomial in the highest derivative by eliminating radicals and fraction powers of the derivatives is called the degree of the differential equation.

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 7 Definite Integrals to solve questions creatively.

Intermediate 2nd Year Maths 2B Definite Integrals Formulas

→ If f is a function integrable over an interval [a, b] and F is a primitive offon [a, b] then
\(\int_{a}^{b}\)f(x)dx = F(b) – F(a)

→ If a < b be real numbers and y = f(x) denote a curve in the plane as shown in figure. Then the definite integral \(\int_{a}^{b}\) f(x) dx is equal to the area of the region bounded by the curve y = f(x), the ordinates x = a, x = b and the portion of X-axis.

→ \(\int_{a}^{b}\)f(x) dx = — \(\int_{b}^{a}\)f(x) dx

→ \(\int_{a}^{b}\) f(x) dx = \(\int_{a}^{c}\) f(x) dx + \(\int_{c}^{b}\)f(x) dx ; a < c < b

→ \(\int_{a}^{b}\)f[g(x)] g’ (x) dx = \(\int_{g(a)}^{g(b)}\)f(x)dx

→ \(\int_{0}^{a}\)f(x)dx = \(\int_{0}^{a}\)f(a-x)dx

→ \(\int_{0}^{2 a}\)f(x)dx =2\(\int_{0}^{a}\)f(x)dx , if f(2a-x) = f(x)
= 0, if f(2a – x) = -f(x)

→ \(\int_{-a}^{a}\)f(x)dx = 2\(\int_{0}^{a}\)f(x)dx if f is even
= 0, if f is odd

→ Let f(x) be a function defined on [a, b]. If ∫ f(x)dx = F(x), then F(b) – F(a) is called the definite integral of f(x) over [a, b]. It is denoted by \(\int_{a}^{b}\) f(x)dx. The real number ‘a’ is called the lower limit and the real number ‘b’ is called the upper limit.
This is known as fundamental theorem of integral calculus.

Geometrical Interpretation of Definite Integral:
If f(x)>0 for all x in [a, b] then \(\int_{a}^{b}\) f (x)dx is numerically equal to the area bounded by the curve y =f(x), the x-axis and the lines x = a and x = b i.e., \(\int_{a}^{b}\) f (x) dx

Properties of Definite Integrals:

• \(\int_{a}^{b}\)f (x)dx = \(\int_{a}^{b}\)f(t)dt i.e., definite integral is independent of its variable.
• \(\int_{a}^{b}\)f(x) dx = – \(\int_{b}^{a}\)f(x) dx
• \(\int_{a}^{b}\) f(x) dx = \(\int_{a}^{c}\) f(x) dx + \(\int_{c}^{b}\)f(x) dx ; a < c < b
• \(\int_{a}^{b}\)f[g(x)] g’ (x) dx = \(\int_{g(a)}^{g(b)}\)f(x)dx
• \(\int_{0}^{a}\)f(x)dx = \(\int_{0}^{a}\)f(a-x)dx
• \(\int_{0}^{2 a}\)f(x)dx =2\(\int_{0}^{a}\)f(x)dx , if f(2a-x) = f(x)
  = 0, if f(2a – x) = -f(x)
• \(\int_{-a}^{a}\)f(x)dx = 2\(\int_{0}^{a}\)f(x)dx if f is even
  = 0, if f is odd

Theorem:
If f(x) is an integrable function on [a, b] and g(x) is derivable on [a, b] then
\(\int_{a}^{b}\)(fog)(x)g'(x)dx = \(\int_{g(a)}^{g(b)}\)f(x) dx \(\int_{a}^{b}\)(fog)(x)g'(x)dx = \(\int_{g(a)}^{g(b)}\)f(x) dx

Areas Under Curves:
1. Let f be a continuous curve over [a, b]. If f(x) ≥ o in [a, b], then the area of the region bounded by y = f(x), x-axis and the lines x=a and x=b is given by \(\int_{a}^{b}\)f(x)dx.

2. Let f be a continuous curve over [a, b]. If f (x) ≤ o in [a, b], then the area of the region bounded by y = f(x), x-axis and the lines x=a and x=b is given by –\(\int_{a}^{b}\) f (x)dx.

3. Let f be a continuous curve over [a,b]. If f (x) ≥ o in [a, c] and f (x) ≤ o in [c, b] where a < c < b. Then the area of the region bounded by the curve y = f(x), the x-axis and the lines x=a and x=b is given by \(\int_{a}^{c}\)f (x )dx – \(\int_{c}^{b}\)f (x )dx

4. Let f(x) and g(x) be two continuous functions over [a, b]. Then the area of the region bounded by the curves y = f (x), y = g (x) and the lines x = a, x=b is given by |\(\int_{a}^{b}\)f(x)dx – \(\int_{a}^{b}\)g(x)dx|

5. Let f(x) and g(x) be two continuous functions over [a, b] and c ∈ (a, b). If f (x) > g (x) in (a, c) and f (x) < g (x) in (c, b) then the area of the region bounded by the curves y = f(x) and y= g(x) and the lines x=a, x=b is given by |\(\int_{a}^{c}\)(f (x) – g (x ))dx| + |\(\int_{c}^{b}\)(g(x)-f (x))dx|

6. let f(x) and g(x) be two continuous functions over [a, bi and these two curves are intersecting at X =x₁ and x = x₂ where x₁, x ₂, ∈ (a,b) then the area of the region bounded by the curves y= f(x) and y = g(x) and the lines x = x₁, x = x₂ is given by |\(\int_{x_{1}}^{x_{2}}\)(f(x) – g(x))dx|

Note: The area of the region bounded by x =g(y) where g is non negative continuous function in [c, d], the y axis and the lines y = c and y = d is given by \(\int_{c}^{d}\)g(y)dy .

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 6 Integration to solve questions creatively.

Intermediate 2nd Year Maths 2B Integration Formulas

→ Integration is the inverse process of differentiation.

→ Let A ⊆ R and let f: A → R be a function. If there is a function B on A such that F'(x) = f(x), ∀ x ∈ A, then we call B an antiderivative of for a primitive of f.
i.e., \(\frac{d}{d x}\)(sin x) = cos x, ∀ x ∈ R ax
f(x) = cos x, x ∈ R, then the function
F(x) = sin x, x ∈ R is an antiderivative or primitive of f.

→ If F is an antiderivative of A, then for k ∈ R, we have (F + k) (x) = f(x), ∀ x ∈ A.

→ Hence F + k is also an antiderivative off.
∴ c is any real number F(x + c) = G(x) = sin x + c, ∀ x ∈ R is also an antiderivative of cos x.

→ It is denoted by ∫ (cos x) dx = sin x + c, (i.e.) ∫ f(x) dx = F(x) + c.

→ Here c is called a constant of integration,
f is called the integrand and x is called the variable of integration.

Standard Forms:

→ ∫xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + c if n ≠ – 1

→ ∫\(\frac{1}{x}\) dx = log |x| + c

→ ∫ sin x dx = – cos x + c, x ∈ R

→ ∫ cos x dx – sin x + c, x ∈ R

→ ∫tan x dx = log |sec x| + c

→ ∫ cot x dx = log |sin x | + c

→ ∫sec x dx = log |sec x + tan x | + c (or) log |tan \(\left(\frac{\pi}{4}+\frac{x}{2}\right)\)| + c

→ ∫cosec x dx = log |cosec x – cot x| + c (or) log |tan\(\left(\frac{x}{2}\right)\)| + c

→ ∫sec² x dx = tan x + c, x ∈ R – \(\left|\frac{n \pi}{2}\right|\), n is odd integer

→ ∫cosec² x dx = – cot x + c → R – nπ, n ∈ Z

→ ∫sec x tan x dx = sec x + c, R – \(\left[\frac{n \pi}{2}\right]\), n is an odd integer

→ ∫cosec x cot xdx = – cosec x + c, R – [nπ], n ∈ Z

→ ∫e^x dx = e^x + c, x ∈ R

→ ∫a^x dx = \(\frac{a^{x}}{\log _{e} a}\) + c, a > 0, a ≠ 1

→ ∫\(\frac{1}{\sqrt{1+x^{2}}}\) dx = sin^-1x + C = – cos^-1 (x) + c

→ ∫\(\frac{d x}{1+x^{2}}\) dx = tan^-1x + C = – cot^-1 (x) + c

→ ∫\(\frac{d x}{|x| \sqrt{x^{2}-1}}\) dx = sec^-1x + C = – cosec^-1 (x) + c

→ ∫ sinh x dx = cosh x + c

→ ∫cosh xdx = sinh x + c

→ ∫cosec²h x dx coth x + c

→ ∫sec²h x dx = tanh x + c

→ ∫cosech x coth xdx = – cosech x + c

→ ∫sech x tanh x dx = – sech x + c

→ ∫e^ax dx = \(\frac{e^{a x}}{a}\) + c

→ ∫e^ax+b dx = \(\frac{e^{a x+b}}{a}\) + c

→ ∫sin (ax + b) dx = \frac{-\cos (a x+b)}{a}\(\) + c

→ ∫cos (ax + b) dx = \(\frac{\sin (a x+b)}{a}\) + c

→ ∫sec² (ax + b) dx = \(\frac{\tan (a x+b)}{a}\) + c

→ ∫cosec² (ax + b) dx = \(\frac{-\cot (a x+b)}{a}\) + c

→ ∫cosec(ax + b) cot(ax + b) dx = \(\frac{-{cosec}(a x+b)}{a}\) + c

→ ∫sec (ax + b) tan(ax + b) dx = \(\frac{\sec (a x+b)}{a}\) + c

→ ∫f(x).g(x) dx = f(x) ∫g(x) dx – ∫[\(\frac{d}{d x}\) f(x) . ∫ g(x) dx] dx (called as integration by parts)

→ ∫\(\frac{1}{\sqrt{1+x^{2}}}\) dx = sinh^-1 (x) + c, x ∈ R = log (x + \(\sqrt{x^{2}+1}\)) + c, x ∈ R

→ ∫\(\frac{1}{\sqrt{x^{2}-1}}\) dx = cosh^-1 (x) + c (or) log (x + \(\sqrt{x^{2}-1}\)) + c, x ∈ (1, ∞)
= – cos h^-1 (- x) + c (or) log (x + \(\sqrt{x^{2}-1}\)) + c, x ∈ (1, ∞)
= log |x + \(\sqrt{x^{2}-1}\)| + c, x ∈ R – [- 1, 1]

→ ∫e^x [f(x) + f'(x)] dx = e^x. f(x) + c

→ ∫\(\frac{1}{\sqrt{a^{2}-x^{2}}}\) dx = sin^-1\(\left(\frac{x}{a}\right)\) + c

→ ∫\(\frac{1}{\sqrt{a^{2}+x^{2}}}\) dx = sinh^-1\(\left(\frac{x}{a}\right)\) + c (or) log \(\frac{\left|x+\sqrt{x^{2}+a^{2}}\right|}{a}\) + c

→ ∫\(\frac{1}{\sqrt{x^{2}-a^{2}}}\) dx = cosh^-1\(\left(\frac{x}{a}\right)\) + c (or) log \(\frac{\left|x+\sqrt{x^{2}-a^{2}}\right|}{a}\) + c

→ ∫\(\frac{1}{a^{2}+x^{2}}\) dx = \(\frac{1}{a}\) tan^-1\(\left(\frac{x}{a}\right)\)+ c

→ ∫\(\frac{1}{\sqrt{a^{2}-x^{2}}}\) = \(\frac{1}{2a}\) log \(\left|\frac{a+x}{a-x}\right|\) + C

→ ∫\(\frac{1}{x^{2}-a^{2}}\) dx = \(\frac{1}{2a}\) log \(\left|\frac{x-a}{x+a}\right|\) + c

→ ∫\(\sqrt{a^{2}-x^{2}}\) dx = \(\frac{1}{2}\)x \(\sqrt{a^{2}-x^{2}}\) + \(\frac{a^{2}}{2}\) sin^-1\(\left(\frac{x}{a}\right)\) + c

→ ∫\(\sqrt{x^{2}+a^{2}}\) dx = \(\frac{1}{2}\)x \(\sqrt{x^{2}+a^{2}}\) + \(\frac{a^{2}}{2}\) sinh^-1\(\left(\frac{x}{a}\right)\) + c

→ ∫\(\sqrt{x^{2}-a^{2}}\) dx = \(\frac{1}{2}\)x \(\sqrt{x^{2}-a^{2}}\) + \(\frac{a^{2}}{2}\) cosh^-1\(\left(\frac{x}{a}\right)\) + c

→ To evaluate

• \(\frac{p x+q}{a x^{2}+b x+c}\) dx
• ∫ (px + q) \(\sqrt{a x^{2}+b x+c}\) dx
• ∫\(\frac{p x+q}{\sqrt{a x^{2}+b x+c}}\) dx, where a, b, c, p, q ∈ R write
  px + q = A.\(\frac{d}{d x}\) (ax² + bx + c) + B and then integrate.

→ To evaluate ∫\(\frac{d x}{(a x+b) \sqrt{p x+q}}\) where a, b, c, p, q, ∈ R put t² = px + q

→ To evaluate ∫\(\frac{1}{a+b \cos x}\) dx (or) \(\frac{1}{a+b \sin x}\) dx
(or) \(\frac{1}{a+b \cos x+c \sin x}\) dx, put tan \(\frac{x}{2}\) = t
Then sin x = \(\frac{2 t}{1+t^{2}}\), cos x = \(\frac{1-t^{2}}{1+t^{2}}\) and dx = \(\frac{2}{1+t^{2}}\) dt

→ To evaluate ∫\(\frac{a \cos x+b \sin x+c}{d \cos x+e \sin x+f}\) dx where a, b, c, d e, f ∈ R; d ≠ 0, e ≠ 0, write a cos x + b sin x + c = A [d cos x + e sin x + f]’ + B (d cos x + e sin x + f) + ∨.
Find A, B, ∨ and then integrate.

→ If I_n = ∫xⁿ . e^ax dx then I_n = \(\frac{x^{n} \cdot e^{a x}}{a}-\frac{n}{a}\) I_{n – 1} for n ∈ N

→ If I_n = ∫ sinⁿ (x) dx then I_n = – \(\frac{\sin ^{n-1}(x) \cos x}{n}\) + \(\left(\frac{n-1}{n}\right)\) I_{n – 2} for n ∈ N, n ≥ 2

→ f I_n = ∫ cosⁿ (x) dx then I_n = – \(\frac{\cos ^{n-1}(x) \sin x}{n}\) + \(\left(\frac{n-1}{n}\right)\) I_{n – 2} for n ∈ N, n ≥ 2

→ If I_n = ∫tanⁿ (x) dx then I_n = \(\frac{\tan ^{n-1}(x)}{n-1}\) I_{n – 2} for N ∈ n, n ≥ 2

→ If I_{m, n} = ∫ sin^m (x) cosⁿ (x) dx then
If I_{m, n} = \(\frac{1}{m+n}\) cos^{n – 1} (x) sin^{m + 1} (x) + \(\left(\frac{n-1}{m+n}\right)\) I_{m, n – 2} where m, n ∈ N, n ≥ 2

→ If I_{m, n} = ∫secⁿ (x) dx then I_n = \(\frac{\sec ^{n-2}(x) \tan x}{n-1}\) + \(\left(\frac{n-2}{n-1}\right)\) I_{n – 2}

Theorem: If f(x) and g(x) are two integrable functions then
∫ f(x).g(x)dx = f(x)∫g(x)dx – ∫f’(x)[∫g(x)dx] dx.
Proof:
\(\frac{\mathrm{d}}{\mathrm{dx}}\) [f(x). ∫g(x)dx] = f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)[∫ g(x)dx] + ∫g(x)dx .\(\frac{\mathrm{d}}{\mathrm{dx}}\)[f(x)]
= f(x)g(x) + [∫g(x)dx]f’(x)
∴ ∫[f(x)g(x) + f’(x)∫g(x)dx] dx = f(x)∫g(x)dx
⇒ ∫f (x)g(x)dx + ∫f’(x) [∫g(x)dx] dx = f (x)∫g(x) dx
∴ ∫f(x)g(x)dx = f(x)∫g(x)dx – ∫f’(x)[∫g(x)dx]dx

Note 1: If u and v are two functions of x then ∫u dv = uv – ∫v du.

Note 2: If u and v are two functions of x; u’, u”, u”’ …………. denote the successive derivatives of u and v₁, v₂, v₃, v₄, v₅ … the successive integrals of v then the extension of integration by pairs is
∫uv dx = uv₁ – u’v₂ + u”v₃ – u”’v₄ + ………

Note 3: In integration by parts, the first function will be taken as the following order.
Inverse functions, Logarithmic functions, Algebraic functions, Trigonometric functions and Exponential functions. (To remember this a phrase ILATE).

Theorem: ∫e^ax cos bx dx = \(\frac{e^{a x}}{a^{2}+b^{2}}\) (a cos bx + b sin bx) + c
Proof:

Theorem: ∫e^ax sin bx dx = \(\frac{e^{a x}}{a^{2}+b^{2}}\) (a sin bx – b cos bx)
Proof:
Let I = ∫e^ax sin bx dx = sin bx ∫e^ax dx – ∫[d(sin bx) ∫e^ax dx] dx

Theorem: ∫ e^x [f(x) + f’(x)]dx = e^xf(x) + c
Proof:
∫e^x [f(x) + f’(x)]dx = ∫e^x f(x)dx + ∫e^x f’(x)dx
= f(x) ∫ e^xdx – ∫[d[f(x)] ∫e^xdx] dx + ∫e^x f'(x)dx
= f(x)e^x – ∫f'(x)e^xdx + ∫e^xf'(x) dx = e^xf(x) + c
Note: ∫e^-x [f(x) – f’(x)]dx = – e^-xf(x) + c

Definition: If f(x) and g(x) are two functions such that f’(x) = g(x) then f(x) is called antiderivative or primitive of g(x) with respect to x.

Note 1: If f(x) is an antiderivative of g(x) then f(x) + c is also an antiderivative of g(x) for all c ∈ R.

Definition: If F(x) is an antiderivative of f(x) then F(x) + c, c ∈ R is called indeVinite integral of f(x) with respect to x. It is denoted by ∫f(x)dx. The real number c s called constant of integration.

Note:

• The integral of a function need not exists. If a function f(x) integral then f(x) is called an integrable function.
• The process of finding the integral of a function is known as Integration.
• The integration is the reverse process of differentiation.

Corollary:
If f(x), g(x) are two integrable functions then ∫(f ± g) (x) dx = ∫f(x)dx ± ∫fg(x)dx

Corollary:
If f₁(x), f₂(x), ……, f_n(x) are integrable functions then
∫(f₁ + f₂ + …….. + f_n)(x)dx = ∫f₁(x)dx + ∫f₂(x)dx + ……. + ∫f_n(x)dx.

Corollary:
If f(x), g(x) are two integrable functions and k, l are two real numbers then ∫(kf + lg) (x)dx = k∫f(x) dx + 1∫g(x)dx.

Theorem: If f f(x)dx = g(x) and a ≠ 0 then ∫ f(ax + b)dx = \(\frac{1}{a}\)g(ax+b)+c.
Proof:
Put ax + b = t.

Theorem: It f(x) is a differentiable function then ∫\(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + c.
Proof:
Put f(x) = t ⇒ f’(x) = \(\frac{d \mathrm{t}}{\mathrm{dx}}\) ⇒ f’(x)dx = dt
∴ ∫\(\frac{f^{\prime}(x)}{f(x)}\) = ∫latex]\frac{1}{\mathrm{t}}[/latex] dt = log |t| + c = log |f(x)| + c

Theorem: ∫tan x dx = log |sec x| for x ≠ (2n + 1)\(\frac{\pi}{2}\), n ∈ Z.
Proof:
∫tan x dx = ∫\(\frac{\sin x}{\cos x}\) dx = -∫\(\frac{d(\cos x)}{\cos x}\) dx
= – log |cos x| + c = log\(\frac{1}{|\cos x|}\) + c = log|sec x| + c

Theorem: ∫cot x dx = log |sin x| + c for x ≠ nπ, n ∈ Z.
Proof:
∫cot x dx = ∫\(\frac{\cos x}{\sin x}\) dx = log |sin x| + c

Theorem: ∫ sec x dx = log |sec x + tan x| + c = log |tan(π/4 + x/2) + c for x ≠ (2n + 1)\(\frac{\pi}{2}\), n ∈ Z.
Proof:

Theorem: ∫csc x dx = log|csc x – cot x| + c = log |tan x/2| + c for x ≠ nπ, n ∈ Z.
Proof:
∫csc x dx = \(\int \frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}\) dx
= \(\int \frac{\csc ^{2} x-\csc x \cot x}{\csc x-\cot x}\) dx = log |csc x – cot x| + c
= log\(\left|\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right|\) + c
= log\(\left|\frac{1-\cos x}{\sin x}\right|\) + c
= log\(\left|\frac{2 \sin ^{2} x / 2}{2 \sin x / 2 \cos x / 2}\right|\) + c
= log |tan x/2| + c

Theorem: If f(x) is differentiable function and n ≠ – 1 then ∫[f(x)]ⁿ f’(x)dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + c.
Proof:
Put f(x) = t ⇒ f’(x) dx = dt

Theorem: If ∫f(x)dx = F(x) and g(x) is a differentiable function then ∫ (fog)(x)g’(x) dx = F[g(x)] + c.
Proof:
g(x) = t ⇒ g’(x) dx = dt
∴ ∫(fog)(x)g’(x)dx = ∫f[g(x)]g’(x) dx
= ∫f(t)dt = F(t) + c = F[g(x)] + c

Theorem: ∫\(\frac{1}{\sqrt{a^{2}-x^{2}}}\) dx = Sin^-1\(\) + c for x ∈ (- a, a)
Proof:
Put x = a sin θ. Then dx = a cos θ dθ

Theorem: ∫\(\frac{1}{\sqrt{a^{2}+x^{2}}}\)dx = Sinh^-1 \(\) + c for x ∈ R.
Proof:
Put x = a sinhθ. Then dx = a cos hθ dθ
∴ ∫\(\frac{1}{\sqrt{a^{2}+x^{2}}}\) dx = \(\int \frac{1}{\sqrt{a^{2}+a^{2} \sinh ^{2} \theta}}\) a coshθ dθ
= ∫\(\frac{a \cosh \theta}{a \cosh \theta}\) = ∫dθ = θ + c = Sinh^-1\(\left(\frac{x}{a}\right)\) + c

Theorem:
∫\(\)dx = Cosh^-1\(\) + c for x ∈ (- ∞, – a) ∪ (a, ∞)
Proof:
Put x = a coshθ. Then dx = a sin hθ dθ
∴ ∫\(\frac{1}{\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}}\) dx = ∫\(\frac{1}{\sqrt{a^{2} \cosh ^{2} \theta-a^{2}}}\) a sin hθ dθ
= ∫\(\frac{a \sinh \theta}{a \sinh \theta}\) dθ = ∫ dθ = θ + c = Cosh^-1\(\left(\frac{x}{a}\right)\) + c

Theorem:
∫\(\frac{1}{a^{2}+x^{2}}\) dx = \(\frac{1}{a}\) Tan^-1\(\left(\frac{x}{a}\right)\) + c for x ∈ R.
Proof:
Put x = a tan θ. Then dx = a sec²θ dθ

Theorem:
∫\(\frac{1}{a^{2}-x^{2}}\)dx = \(\frac{1}{2 \mathrm{a}}\) log\(\left|\frac{a+x}{a-x}\right|\) + c for x ≠ ± a
Proof:
∫\(\frac{1}{a^{2}-x^{2}}\)dx = ∫\(\left|\frac{a+x}{a-x}\right|\)dx
= \(\frac{1}{2 \mathrm{a}} \int\left(\frac{1}{\mathrm{a}+\mathrm{x}}+\frac{1}{\mathrm{a}-\mathrm{x}}\right)\)dx = \(\frac{1}{2 \mathrm{a}}\) [log |a + x| – log |a – x|] + c
= \(\frac{1}{2 \mathrm{a}}\) log\(\left|\frac{a+x}{a-x}\right|\) + c

Theorem:
∫\(\frac{1}{x^{2}-a^{2}}\) dx = \(\frac{1}{2 \mathrm{a}}\) log \(\left|\frac{x-a}{x+a}\right|\) + c for x ≠± a
Proof:
∫\(\frac{1}{x^{2}-a^{2}}\) dx = ∫\(\frac{1}{(x-a)(x+a)}\) dx
= \(\frac{1}{2 a} \int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)\) dx = \(\frac{1}{2 \mathrm{a}}\) [log |x – a| – log |x + a|] + c
= \(\frac{1}{2 \mathrm{a}}\) log \(\left|\frac{x-a}{x+a}\right|\) + c

Theorem:
∫\(\sqrt{a^{2}-x^{2}}\)dx = \(\frac{x}{2} \sqrt{a^{2}-x^{2}}\) + \(\frac{\mathrm{a}^{2}}{2}\) sin^-1\(\left(\frac{x}{a}\right)\) + c for x ∈ (- a, a)
Proof:
Put x = a sin θ. Then dx = a cos θ dθ

Theorem:
∫\(\sqrt{a^{2}+x^{2}}\) dx = \(\frac{x}{2} \sqrt{a^{2}+x^{2}}\) + \(\frac{\mathrm{a}^{2}}{2}\) Sinh^-1 \(\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\) + c for x ∈ R.
Proof:
Put x = sinhθ. Then dx = a coshθ dθ
∴ ∫\(\sqrt{a^{2}+x^{2}}\) dx = ∫\(\sqrt{a^{2}+a^{2} \sinh ^{2} \theta}\) a coshθ dθ
= ∫\(a \sqrt{1+\sinh ^{2} \theta}\) a coshθ dθ = a² ∫cosh² θdθ
= \(=\mathrm{a}^{2} \int \frac{1+\cosh 2 \theta}{2} \mathrm{~d} \theta=\frac{\mathrm{a}^{2}}{2}\left[\theta+\frac{1}{2} \sinh 2 \theta\right]+\mathrm{c}\)
= \(\frac{a^{2}}{2}\left[\theta+\frac{1}{2} 2 \sinh \theta \cosh \theta\right]+c\)
= \(\frac{a^{2}}{2}\left[\theta+\sinh \theta \sqrt{1+\sinh ^{2} \theta}\right]+c\)
= \(\frac{a^{2}}{2}\left[{Sinh}^{-1}\left(\frac{x}{a}\right)+\frac{x}{a} \sqrt{1+\frac{x^{2}}{a^{2}}}\right]+c\)
= \(\frac{\mathrm{a}^{2}}{2}{Sinh}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\frac{\mathrm{x}}{\mathrm{a}} \sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}+\mathrm{c}\)

Theorem:
∫\(\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}\) dx = \(\frac{x}{2} \sqrt{x^{2}-a^{2}}\) – \(\frac{a^{2}}{2}\) Cosh^-1\(\left(\frac{x}{a}\right)\) + c for x ∈ [a, ∞)
Proof:
Put x = a coshθ. Then dx = a sinhθ dθ

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 5 Hyperbola to solve questions creatively.

Intermediate 2nd Year Maths 2B Hyperbola Formulas

Definition:
A conic with an eccentricity greater than one is called a hyperbola, i.e., the locus of a point, the ratio of whose distances from a fixed point (focus) and fixed straight line (directrix) is e, where e> 1 is called a hyperbola.

Equation of hyperbola in standard form:

→ \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
Here b² = a² (e² – 1)

→ The difference of the focal distances of any point on the hyperbola is constant.

(i.e.,) SP ~ S’P = 2a

→ A Point P(x₁, y₁) in the plane of the hyperbola S = 0 lies inside the hyperbola, if S₁₁ < 0, lies outside if S₁₁ > 0 and on the curve S₁₁ = 0

→ Ends of the latus rectum (± ae, ± b²/a) and the length of the latus rectum is 2b²/a.

→ Equation of tangent at P(x₁, y₁) to \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
\(\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}\) = 1

→ Equation of normal at P(x₁, y₁)
\(\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}\) = a² + b²

→ Parametric form is x = a sec θ, y = b tan θ.

→ Equation of tangent at ‘θ’ on the hyperbola
\(\frac{x}{a}\) sec θ – \(\frac{y}{b}\) tan θ = 1

→ Equation of normal at θ on the hyperbola
\(\frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}\) = a² + b²

→ A necessary and sufficient condition for a straight line y = mx + c to be tangent to hyperbola c² = a²m² – b²
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

→ If P(x₁, y₁) is any point in the plane of the hyperbola S = 0 then the equation of the polar of P is S₁ = 0.

→ The pole of the line lx + my + n = 0; n ≠ 0 with respect to the hyperbola S = 0 is
\(\left(\frac{-a^{2} l}{n} ; \frac{b^{2} m}{n}\right)\), n ≠ 0

Equation of a Hyperbola in Standard From.
The equation of a hyperbola in the standard form is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Proof:
Let S be the focus, e be the eccentricity and L = 0 be the directrix of the hyperbola.
Let P be a point on the hyperbola. Let M, Z be the projections of P, S on the directrix L = 0 respectively. Let N be the projection of P on SZ. Since e > 1, we can divide SZ both internally and externally in the ratio e : 1.

Let A, A’ be the points of division of SZ in the ratio e : 1 internally and externally respectively. Let AA’ = 2a. Let C be the midpoint of AA’. The points A, A’ lie on the hyperbola and \(\frac{\mathrm{SA}}{\mathrm{AZ}}\) = e, \(\frac{\mathrm{SA}^{\prime}}{\mathrm{A}^{\prime} \mathrm{Z}}\) = e

∴ SA = eAZ, SA’ = eA’Z
Now SA + SA’ = eAZ + eA’Z
⇒ CS – CA + CS + CA’ = e(AZ + A’Z)
⇒ 2CS = eAA’ (∵ CA = CA’)
⇒ 2CS = e2a ^ CS = ae
Aslo SA’ – SA = eA’Z – eAZ
AA’ = e(A’Z – AZ)
⇒ 2a = e[CA’ + CZ – (CA – CZ)]
⇒ 2a = e 2CZ (∵ CA = CA’) ⇒ CZ = \(\frac{a}{e}\).

Take CS, the principal axis of the hyperbola as
x-axis and Cy perpendicular to CS as y-axis. Then S = (ae, 0).
Let P(x₁, y₁).
Now PM = NZ = CN – CZ = x₁ – \(\frac{a}{e}\).
P lies on the hyperbola ⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = e
⇒ PS = ePM ⇒ PS² = e²PM²
⇒ (x₁ – ae)² + (y₁ – 0)² = e² \(\left(x_{1}-\frac{a}{e}\right)^{2}\)
⇒ (x₁ – ae)² + y₁² = (x₁e – a)²
⇒ x₁² + a² e² – 2x₁ae + y₁² = x₁e² + a² – 2x₁ae
⇒ x²(e² -1) – y² = a²(e² -1)
⇒ \(\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{a^{2}\left(e^{2}-1\right)}\) = 1 ⇒ \(\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}\) = 1
Where b² = a²(e² – 1)
The locus of P is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
The equation of the hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Nature of the curve \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

Let C be the curve represented by \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1. Then
(i) (x,y) ∈ C(x, -y) ∈ C and (x, y) ∈ C ⇔ (-x,y) ∈ C.
Thus the curve is symmetric with respect to both the x-axis and the y-axis. Hence the coordinate axes are two axes of the hyperbola.

(ii) (x, y) ∈ C ⇔ (-x, -y) ∈ C.
Thus the curve is symmetric about the origin O and hence O is the midpoint of every chord of the hyperbola through O. Therefore the origin is the center of the hyperbola.

(iii) (x, y) ∈ C and y = 0 ⇒ x² = a² ⇒ x = ±a.
Thus the curve meets x-axis (Principal axis) at two points A(a, 0), A'(-a, 0). Hence hyperbola has two vertices. The axis AA’ is called transverse axis. The length of transverse axis is AA’ = 2a.

(iv) (x, y) ∈ C and x = 0 ⇒ y² = -b² ⇒ y is imaginary.
Thus the curve does not meet the y-axis. The points B(0, b), B'(0, -b) are two points on y-axis. The axis BB’ is called conjugate axis. BB’ = 2b is called the length of conjugate axis.

(v) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 ⇒ y = \(\frac{b}{a} \sqrt{x^{2}-a^{2}}\) ⇒ y has no real value for -a < x < a.
Thus the curve does not lie between x = -a and x = a.
Further x → ∞ ⇒ y ^ ± ∞ and
x → -∞ ⇒ y → ± ∞.
Thus the curve is not bounded (closed) on both the sides of the axes.

(vi) The focus of the hyperbola is S(ae, 0). The image of S with respect to the conjugate axis is S'(-ae, 0). The point S’ is called second focus of the hyperbola.

(vii) The directrix of the hyperbola is x = a/e. The image of x = a/e with respect to the conjugate axis is x = -a/e. The line x = -a/e is called second directrix of the hyperbola corresponding to the second focus S’.

Theorem:
The length of the latus rectum of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{2 b^{2}}{a}\)
Proof:
Let LL be the length of the latus rectum of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

If SL = 1, then L = (ae, 1)
L lies on the hyperbola ⇒ \(\frac{(a \mathrm{e})^{2}}{a^{2}}-\frac{1^{2}}{b^{2}}\) = 1
⇒ \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = e² – 1 ⇒ 1² = b²(e² – 1)
⇒ 1 = b² × \(\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\) ⇒ 1 = \(\frac{b^{2}}{a}\) ⇒ SL = \(\frac{b^{2}}{a}\)
∴ LL’ = 2SL = \(\frac{2 b^{2}}{a}\)

Theorem:
The difference of the focal distances of any point on the hyperbola is constant f P is appoint on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 with foci S and S then |PS’- PS| = 2a
Proof:
Let e be the eccentricity and L = 0, L’ = 0 be the directrices of the hyperbola.
Let C be the centre and A, A’ be the vertices of the hyperbola.
∴ AA’ = 2a

Foci of the hyperbola are S(ae, 0), S'(-ae, 0).
Let P(x₁, y₁) be a point on the hyperbola.
Let M, M’ be the projections of P on the directrices L = 0, L’ = 0 respectively.
∴ \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e, \(\frac{\mathrm{S}^{\prime} \mathrm{P}}{\mathrm{PM}^{\prime}}\) = e

Let Z, Z’ be the points of intersection of transverse axis with directrices.
∴ MM’ = ZZ’ = CZ + CZ’ = 2a/e
PS’ – PS = ePM’ – ePM = e(PM’ – PM)
= e(MM’) = e(2a/e) = 2a

Notation: We use the following notation in this chapter.
S ≡ \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) – 1, S₁ = \(\frac{x_{1}}{a^{2}}-\frac{y_{1}}{b^{2}}\) – 1
S₁₁ ≡ S(x1, y1) = \(\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}\) – 1, S₁₂ = \(\frac{\mathrm{x}_{1} \mathrm{x}_{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}_{1} \mathrm{y}_{2}}{\mathrm{~b}^{2}}\) – 1

Note: Let F(x_1, y₁) be a point and S ≡ \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) – 1 = 0 be a hyperbola. Then

• F lies on the hyperbola S = 0 ⇔ S₁₁ = 0
• F lies inside the hyperbola S = 0 ⇔ S₁₁ > 0
• F lies outside the hyperbola S = 0 ⇔ S₁₁ < 0

Theorem:
The equation of the chord joining the two points A(x₁, y₁), B(x₂, y₂) on the hyperbola S = 0 is S₁ + S₂ = S₁₂.

Theorem:
The equation of the normal to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 at P(x₁ y₁) is \(\frac{\mathrm{a}^{2} \mathrm{x}}{\mathrm{x}_{1}}+\frac{\mathrm{b}^{2} \mathrm{y}}{\mathrm{y}_{1}}\) = a² + b².

Theorem: The condition that the line y = mx + c may be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is c² = a²m² – b²

Note: The equation of the tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 may be taken as y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
The point of contact is \(\left(\frac{-a^{2} m}{c}, \frac{-b^{2}}{c}\right)\) where c = am – b

Theorem: Two tangents can be drawn to a hyperbola from an external point.
Note: If m₁, m₂ are the slopes of the tangents through P, then m₁ m₂ become the roots of (x₂² – a²)m² – 2x₁y₁m + (y₁² + b²) = 0
Hence m₁ + m₂ = \(\frac{2 \mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{x}_{1}^{2}-\mathrm{a}^{2}}\)
m₁m₂ = \(\frac{\mathrm{y}_{1}^{2}+\mathrm{b}^{2}}{\mathrm{x}_{1}^{2}-\mathrm{a}^{2}}\)

Theorem:
The point of intersection of two perpendicular tangents to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 lies on the circle x² + y² = a² – b².
Proof:
Equation of any tangent to the hyperbola is:
y = mx ± \(\)
Suppose P(x₁, y₁) is the point of intersection of tangents.
Plies on the tangent Y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\) y₁ = mx₁ = ±\(\sqrt{a^{2} m^{2}-b^{2}}\)
= (y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ – a²m² + b² = 0
= m²(x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0

This is a quadratic in m giving the values for m say m1 and m2.
The tangents are perpendicular
⇒ m₁m₂ = -1 ⇒ \(\frac{\mathrm{y}_{1}^{2}+\mathrm{b}^{2}}{\mathrm{x}_{1}^{2}-\mathrm{a}^{2}}\) = -1
⇒ y₁² + b² = -x² + a² ⇒ x₁² + y₁² = a² – b²
P(x₁, y₁) lies on the circle x² + y = a² – b².

Definition:
The point of intersection of perpendicular tangents to a hyperbola lies on a circle, concentric with the hyperbola. This circle is called director circle of the hyperbola.

Corollary:
The equation to the auxiliary circle of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\)= 1 is x² + y² = a².

Theorem:
The equation to the chord of contact of P(x₁, y₁) with respect to the hyperbola S = 0 is S₁ = 0.

Midpoint of a Chord:
Theorem: The equation of the chord of the hyperbola S = 0 having P(x₁, y₁) as it’s midpoint is S₁ = S₁₁.

Pair of Tangents:
Theorem: The equation to the pair of tangents to the hyperbola S = 0 from P(x₁, y₁) is S₁² = S₁₁S

Asymptotes:
Definition: The tangents of a hyperbola which touch the hyperbola at infinity are called asymptotes of the hyperbola.

Note:

• The equation to the pair of asymptotes of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 0.
• The equation to the pair of asymptotes and the hyperbola differ by a constant.
• Asymptotes of a hyperbola passes through the centre of the hyperbola.
• Asymptotes are equally inclined to the axes of the hyperbola.
• Any straight line parallel to an asymptote of a hyperbola intersects the hyperbola at only one point.

Theorem:
The angle between the asymptotes of the hyperbola S = 0 is 2tan^-1(b/a).
Proof:
The equations to the asymptotes are y = ± \(\frac{b}{a}\)x.
If θ is an angle between the asymptotes, then
tan θ = \(\frac{\frac{\mathrm{b}}{\mathrm{a}}-\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)}{1+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)}=\frac{2\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}\) = tan 2α Where tan α = \(\frac{b}{a}\)
∴ θ = 2α = 2Tan^-1\(\frac{b}{a}\)

Parametric Equations:
A point (x, y) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 represented as x = a sec θ, y = b tan θ in a single parameter θ. These equations x = a sec0, y= b tan θ are called parametric equations of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
The point (a sec θ, b tan θ) is simply denoted by θ.

Note: A point on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 can also be represented by (a cosh θ, b sinh θ). The equations x = a cosh θ, y = sinh θ are also called parametric equations of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

Theorem: The equation of the chord joining two points α and β on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is:
\(\frac{x}{a}\)cos\(\frac{\alpha-\beta}{2}\) – \(\frac{y}{b}\)sin\(\frac{\alpha+\beta}{2}\) = cos\(\frac{\alpha+\beta}{2}\)

Theorem:
The equation of the tangent at P(0) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{x}{a}\)cos\(sec θ – [latex]\frac{y}{b}\) tan θ = 1.

Theorem:
The equation of the normal at P(0) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{a x}{\sec \theta}+\frac{\text { by }}{\tan \theta}\) = a² + b².

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 4 Ellipse to solve questions creatively.

Intermediate 2nd Year Maths 2B Ellipse Formulas

Definition:
→ A conic with eccentricity less than one is called an ellipse i.e., the locus of a point whose distances from a fixed point and a fixed straight line are in constant ratio ‘e’ which is less than 1, is called an ellipse. The fixed point and fixed straight line are called focus and directrix respectively.

Equation of Ellipse in standard form:
→ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1; a > b, b² = a² (1 – e²) ; e < 1
Foci: (ae, 0), (- ae, 0); directrices x = \(\frac{\mathrm{a}}{\mathrm{e}}\); x = \(\frac{-a}{e}\)
a = Length of the semi-major axis.
b = Length of semi-minor axis.

Various forms of the ellipse:

→ Major axis: along X – axis
Length of major axis: 2a
Minor axis: along Y-axis
Length of minor axis: 2b
Centre: (0, 0)
Foci: S (ae, 0); S’ (- ae, 0)
Directrices: x = \(\frac{\mathrm{a}}{\mathrm{e}}\); x = – \(\frac{\mathrm{a}}{\mathrm{e}}\)
e is given as: b² = a² (1 – e²)
(or) e = \(\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}\), a > b
a < b (or) b > a

→ Major axis: along Y – axis
Length of major axis: 2b
Minor axis: along Y-axis
Length of minor axis: 2a
Centre: (0, 0)
Foci: S (0, be); S’ (0, – be)
Directrices: y = \(\frac{b}{\mathrm{e}}\), y = – \(\frac{b}{\mathrm{e}}\)
e is given as: a² = b² (1 – e²)
(or) e = \(\sqrt{\frac{b^{2}-a^{2}}{b^{2}}} .\)

→ \(\frac{(x-\alpha)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}\) = 1, a > b
Major axis: parallel to X-axis along the line y = β
Length of major axis: 2a
Minor axis: parallel to Y-axis along the line x = α
Length of minor axis: 2b
Center: (α, β)
Foci: S (ae + α, β); S’ ( – ae + α, β)
Directrices: x = α + \(\frac{a}{e}\) ; x = α – \(\frac{a}{e}\)
e is given by b² = a² ( 1 – e²)

→ \(\frac{(x-\alpha)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}\) = 1, a < b
Major axis: parallel to Y-axis along the line x = α
Length of major axis: 2b
Minor axis: parallel to X-axis along the line y = β
Length of minor axis: 2a
Center: (α, β)
Foci: S (α, be + β); S’ (α, -be + β)
Directrices: y – β = \(\frac{b}{\mathrm{e}}\) ; y – β = – \(\frac{b}{\mathrm{e}}\)

→ A line segment joining two points on the ellipse is called a chord of the ellipse. Chord passing through foci is called a focal chord. A focal chord perpendicular to the major axis of the ellipse is cal fed latus rectum. Length of latus rectum = \(\frac{2 b^{2}}{a}\), a > b, Length of the latus rectum = \(\frac{2 a^{2}}{a}\), if b > a

→ If P (x, y) is any point on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 whose foci are S and S’.
Then SP + SP’ = constant = 2a.

→ Parametric equations x = a cos θ; y = b sin θ

→ If P a point lies outside the ellipse, then S₁₁ > 0.

→ If P a point lies on the ellipse, then S₁₁ = 0.

→ If P a point lies inside the ellipse, then S₁₁ < 0.

Ellipse:
A conic is said to be an ellipse if it’s eccentricity e is less than 1.

Equation of an Ellipse in Standard Form:
The equation of an ellipse in the standard form is \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}\) = 1.(a < b)
Proof:

Let S be the focus, e be the eccentricity and L = 0 be the directrix of the ellipse. Let P be a point on the ellipse. Let M, Z be the projections (foot of the perpendiculars) of P, S on the directrix L = 0 respectively. Let N be the projection of P on SZ. Since e < 1, we can divide SZ both internally and externally in the ratio e: 1. Let A, A’ be the points of division of SZ in the ratio e: 1 internally and externally respectively. Let AA’ = 2a. Let C be the midpoint of AA’. The points A, A’ lie on the ellipse and
\(\frac{S A}{A Z}\) = eAZ, \(\frac{\mathrm{SA}^{\prime}}{\mathrm{A}^{\prime} \mathrm{Z}}\) = eA’Z
Now SA + SA’ = eAZ + eA’Z
⇒ AA’ = e(AZ + A’Z)
⇒ 2a = e(CZ – CA + A’C + CZ)
⇒ 2a = e.2CZ (∵ CA = A’C)
⇒ CZ = a/e

Also SA’- SA = eA’Z – eAZ
⇒ A’C + CS – (CA – CS) = e(A’Z – AZ)
⇒ 2CS = eAA’ (∵ CA = A’C)
⇒ 2CS = e2a ⇒ CS = ae
Now PM = NZ = CZ – CN = \(\frac{a}{e}\) – x₁

P lies on the ellipse:
⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = e ⇒ PS = ePM ⇒ PS² = e²PM²
⇒ (x₁ – ae)² + (y₁ – 0)² = e²\(\left(\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{x}_{1}\right)^{2}\)
⇒ (x₁ – ae)² + y₁² = (a – x₁e)²
⇒ x₁ + ae – 2x₁ae + y₁ = a + x₁e – 2x₁ae
⇒ (1 – e²)x₁² + y₁² = (1 – e²)a²
⇒ \(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{a^{2}\left(1-e^{2}\right)}\) = 1 ⇒ \(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}\) = 1
Where b² = a²(1 – e²) > 0
The locus of P is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
∴ The equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Nature of the Curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.

Let be the curve represented by \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1. Then

• The curve is symmetric about the coordinate axes.
• The curve is symmetric about the origin O and hence O is the midpoint of every chord of the ellipse through O. Therefore the origin is the centre of the ellipse.
• Put y = 0 in the equation of the ellipse ⇒ x = a ⇒ x = ±a.
  Thus the curve meets x-axis (Principal axis) at two points A(a, 0), A'(-a, 0). Hence the ellipse has two vertices. The axis AA’ is called major axis. The length of the major axis is AA’ = 2a
• Put x = 0 ⇒ y² = b² ⇒ y = ± b. Thus, the curve meets y-axis (another axis) at two points B(0, b), B'(0, -b). The axis BB’ is called minor axis and the length of the minor axis is BB’ = 2b.
• The focus of the ellipse is S(ae, 0). The image of S with respect to the minor axis is S'(-ae,0). The point S’ is called second focus of the ellipse.
• The directrix of the ellipse is x = a/e. The image of x = a/e with respect to the minor axis is x = -a/e. The line x = -a/e is called second directrix of the ellipse.
• \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
  y² = b²(1 – \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}\)) ⇒ y = \(\frac{b}{a} \sqrt{a^{2}-x^{2}}\)

Thus y has real values only when -a ≤ x ≤ a. Similarly x has real values only when -b ≤ y ≤ b. Thus the curve lies completely with in the rectangle x = ±a, y = ±b. Therefore the ellipse is a closed curve.

Theorem:
The length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (a > b > 0) is \(\frac{2 b^{2}}{a}\)
The length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 (0 < a < b) is \(\frac{2 b^{2}}{a}\)

Let LL’ be the length of the latus rectum of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Focus S = (ae, 0)
If SL = 1, then L = (ae, 1)
L is lies on the ellipse ⇒ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
⇒ e + \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = 1 ⇒ \(\frac{1^{2}}{\mathrm{~b}^{2}}\) = 1 – e = \(\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\) ⇒ 1 = \(\frac{b^{4}}{a^{2}}\)
⇒ 1 = \(\frac{b^{2}}{a}\) ⇒ SL = \(\frac{b^{2}}{a}\)
LL’ = 2SL = \(\frac{2 b^{2}}{a}\)

Note: The coordinates of the four ends of the latus recta of the ellipse \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\) = 1 (a < b < 0) are L = (ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)), L’ = (ae, –\(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)), L₁ = (-ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\)); L₁‘ = (-ae, \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\))

Note: The coordinates of the four ends of the latus recta of the ellipse \(\frac{\mathrm{b}^{2}}{\mathrm{a}}\) = 1 (0 < a < b) are L = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), be), L’ = (-\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), be), L₁ = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), -be), L₁‘ = (\(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\), -be)

Theorem:
If P is a point on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 with foci S and S then PS + PS’ = 2a.
Proof:
Let e be the eccentricity and L = 0, L’ = 0 be the directrices of the ellipse.
Let C be the centre and A, A’ be the vertices of the ellipse.
∴ AA’ = 2a
Foci of the ellipse are S(ae, 0), S'(-ae, 0)
Let P(x₁, y₁) be a point on the ellipse

Let M, M’ be the projections of P on the directrices L = 0, L’ = 0 respectively.
∴ \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e, \(\frac{\mathrm{S}^{\prime} \mathrm{P}}{\mathrm{PM}^{\prime}}\) = e

Let Z, Z’ be the points of intersection of major axis with directrices.
∴ MM’ = ZZ’ = CZ + CZ’ = 2a/e.
PS + PS’ = ePM + ePM’
= e(PM + PM’) = e(MM’) = e(2a/e) = 2a.

Theorem:
Let P(x₁, y₁) be a point and S ≡ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 = 0 be an ellipse. Then
(i) P lies on the ellipse ⇔ S₁₁ = 0,
(ii) P lies inside the ellipse ⇔ S₁₁ < 0,
(iii) P lies outside the ellipse ⇔ S₁₁ > 0

Theorem:
The equation of the tangent to the ellipse S = 0 at F(x₁, y₁) is S₁ = 0.

Theorem:
The equation of the normal to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 F(x₁, y₁) is \(\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}\) = a² – b².
Proof:
The equation of the tangent to S = 0 at F is S₁ = 0
⇒ \(\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{~b}^{2}}\) – 1 =
The equation of the normal to S = 0 at F is
\(\frac{\mathrm{y}_{1}}{\mathrm{~b}^{2}}\)(x – x₁) – \(\frac{\mathrm{x}_{1}}{\mathrm{~a}^{2}}\)(y – y₁) = 0
⇒ \(\frac{x_{1}}{b^{2}}-\frac{y_{1}}{a^{2}}=\frac{x_{1} y_{1}}{b^{2}}-\frac{x_{1} y_{1}}{a^{2}}\)
⇒ \(\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{x}_{1} \mathrm{y}_{1}}\left(\frac{\mathrm{xy}}{\mathrm{b}^{2}}-\frac{\mathrm{y} \mathrm{x}_{1}}{\mathrm{a}^{2}}\right)=\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{x}_{1} \mathrm{y}_{1}}\left(\frac{\mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{~b}^{2}}-\frac{\mathrm{x}_{1} \mathrm{y}_{1}}{\mathrm{a}^{2}}\right)\)
⇒ \(\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}\) = a² – b²

Theorem:
The condition that the line y = mx + c may be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 is c² = a²m² + b².
Proof:
Suppose y = mx + c … (1)is a tangent to the e11ipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1.
Let P(x1, y) be the point of contact.
The equation of the tangent at P is
\(\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{~b}^{2}}\) – 1 = 0 … (2)

Now (1) and (2) represent the same line.
\(\frac{\mathrm{x}_{1}}{\mathrm{a}^{2} \mathrm{~m}}=\frac{\mathrm{y}_{1}}{\mathrm{~b}^{2}(-1)}=\frac{-1}{\mathrm{c}}\) ⇒ x₁ = \(\frac{-a^{2} m}{c}\), y₁ = \(\frac{\mathrm{b}^{2}}{\mathrm{c}}\)
P lies on the line y = mx + c ⇒ y₁ = mx₁ + c
⇒ \(\frac{\mathrm{b}^{2}}{\mathrm{c}}\) = m\(\left(\frac{-a^{2} m}{c}\right)\) + c ⇒ b² = -a²m² + c²
⇒ c² = a²m² + b²

Note:
The equation of a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 may be taken as y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\). The point of contact is \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\) where c² = a²m² + b²

Theorem:
Two tangents can be drawn to an ellipse from an external point.

Director Circle:
The points of intersection of perpendicular tangents to an ellipse S = 0 lies on a circle, concentric with the ellipse.
Proof:
Equation of the ellipse
S ≡\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 0
Let P(x₁, y₁) be the point of intersection of perpendicular tangents drawn to ellipse.

Let y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) be a tangent to the ellipse S = 0 passing through P.
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
⇒ y₁ – mx₁ = ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
⇒ (y₁ – mx₁)² = a²m² + b²
⇒ y₁² + m²x₁² – 2x₁y₁m = a²m² + b²
⇒ (x₁² – a²)m² -2x₁y₁m + (y₁² -b²) = 0 … (1)

If m₁, m₂ are the slopes of the tangents through P then m1, m2 are the roots of (1).
The tangents through P are perpendicular.
⇒ m₁m₂ = -1 ⇒ \(\frac{y_{1}^{2}-b^{2}}{x_{1}^{2}-a^{2}}\) = -1
⇒ y₁² – b² = -x² + a² ⇒ x²₁ + y²₁ = a² + b²
∴ P lies on x² + y² = a² + b² which is a circle with centre as origin, the centre of the ellipse.

Auxiliary Circle:
Theorem: The feet of the perpendiculars drawn from either of the foci to any tangent to the ellipse S = 0 lies on a circle, concentric with the ellipse.( called auxiliary circle)
Proof:
Equation of the ellipse S ≡ \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) – 1 = 0
Let P(x₁, y₁) be the foot of the perpendicular drawn from either of the foci to a tangent.
The equation of the tangent to the ellipse S = 0 is y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\) … (1)
The equation to the perpendicular from either foci (±ae, 0) on this tangent is
y = –\(\frac{1}{m}\)(x ± ae)

Now P is the point of intersection of (1) and (2)

∴ y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\), y₁ = –\(\frac{1}{m}\)(x₁ ± ae
⇒ y₁ – mx₁ = ±V a2m2 + b2, my₁ + x₁ = ±ae
⇒ (y₁ – mx₁)² + (my₁ + x₁)² = a²m² + b² + a²e²
⇒ y₁² + m²x² – 2x₁y₁m + m²y₁² + x₁² + 2x₁y₁m = a²m² + a²(1 – e²) + a²e²
⇒ x₁²(m² + 1) + y₁²(1 + m²) = a²m² + a²
⇒ (x₁² + y₁²)(m² + 1) = a² (m² + 1)
⇒ x₁² + y₁² = a²
P lies on x² + y² = a² which is a circle with centre as origin, the centre of the ellipse.

Theorem:
The equation to the chord of contact of P(x₁, y₁) with respect to the ellipse S = 0 is S₁ = 0.

Eccentric Angle Definition:
Let P(x, y) be a point on the ellipse with centre C. Let N be the foot of the perpendicular of P on the major axis. Let NP meets the auxiliary circle at P’. Then ∠NCP’ is called eccentric angle of P. The point P’ is called the corresponding point of P.

Parametric Equations: If P(x, y) is a point on the ellipse then x = a cos θ, y = b sin θ where θ is the eccentric angle of P. These equations x = a cos θ, y = b sin θ are called parametric equations of the ellipse. The point P(a cos θ, b sin θ) is simply denoted by θ.

Theorem: The equation of the chord joining the points with eccentric angles α and β on the ellipse S = 0 is \(\frac{x}{a} \cos \frac{\alpha+\beta}{2}+\frac{y}{b} \sin \frac{\alpha+\beta}{2}=\cos \frac{\alpha-\beta}{2}\)
Proof:
Given points on the ellipse are P(a cos α, b sin α), Q(a cos β, b sin β).
Slope of \(\overline{\mathrm{PQ}}\) is \(\frac{\mathrm{b} \sin \alpha-\mathrm{b} \sin \beta}{\mathrm{a} \cos \alpha-\mathrm{a} \cos \beta}=\frac{\mathrm{b}(\sin \alpha-\sin \beta)}{\mathrm{a}(\cos \alpha-\cos \beta)}\)
Equation of \(\overline{\mathrm{PQ}}\) is:

Theorem: The equation of the tangent at P(θ) on the ellipse
S = 0 is \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1.

Theorem: The Equation of The Normal At P(θ) On The Ellipse
S = 0 Is \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}\) = a² – b².

Theorem: Four normals can be drawn from any point to the ellipse and the sum of the eccentric angles of their feet is an odd multiple of π.

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 3 Parabola to solve questions creatively.

Intermediate 2nd Year Maths 2B Parabola Formulas

Definition:
→ Conic section: If right circular solid cone is cut by a plane the section of it is called a Conic section.

→ Conic: The locus of a point whose distances from a fixed point and a fixed straight line are in constant ratio ‘e’ is called a conic.

→ Parabola : A conic with eccentricity 7 is called a parabola i.e., the locus of a point, whose distance from fixed point (focus) is equal to the distance from a fixed straight line (directrix) is called a parabola.

→ Axis of Parabola: The line passing through the vertex and the focus and perpendicular to directrix of the parabola is called the axis of the Parabola.

→ Equation of Parabola : General form – Let S(α, β) be the focus and ax + by + c = 0 be directrix then by definition the equation of parabola be,
\(\sqrt{(x-\alpha)^{2}+(y-\beta)^{2}}\) = \(\left|\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\right|\)

Various forms of parabola:
→ y² = 4ax
Axis : X – axis
Focus : (a, 0)
Vertex : (0, 0)
Equation of directrix : x + a = 0

→ y²= – 4ax
Axis: x – axis
Focus: (- a, 0)
Vertex : (0, 0)
Equation of directrix : x – a = 0

→ x² = 4ay (a > 0)
Axis: Y-axis Focus : (0, a)
Vertex : (0, 0)
Equation of directrix: y + a = 0

→ x² = -4ay(a>0)
Axis : Y axis
Focus : (0, – a)
Vertex: (0, 0)
Equation of directrix : y – a = 0

→ (y – k)² = 4a (x – h) a > 0
Axis : y = k’ a line parallel to X- axis
Focus : (a + h, k)
Vertex: (h, k)
Equation of directrix: x + a = h

→ (x – h)² = 4a (y – k) a > 0
Axis: x = h a line parallel to Y-axis
Focus: (h, a + k)
Vertex: (h, k)
Equation of directrix: y + a = k

→ \(\sqrt{(x-h)^{2}+(y-k)^{2}}\) = \(\left|\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\right|\)
Axis: b(x – h) – a(y – k) = 0
Focus: (h, k)
Vertex: Some point A(fig)
Equation of directrix : ax + by + c = 0

Chord: Line segment joining two points of a parabola is called a, chord of the parabola. If this chord passes through focus is called focal chord.
Chord passing through focus and ⊥ to axis is called latus rectum.
Length of latus rectum = 4a

Parametric form of parabola:
For y² = 4ax
Parametric form will be x = at², y = 2at

Definition:
→ The locus of a point which moves in a plane so that its distance from a fixed point bears a constant ratio to its distance from a fixed straight line is called a conic section or conic. The fixed point is called focus, the fixed straight line is called directrix and the constant ratio ‘e’ is called eccentricity of the conic.

If e = 1, then the conic is called a Parabola.
If e < 1, then the conic is called an Ellipse. If e > 1, then the conic is called a hyperbola.
Note: The equation of a conic is of the form ax² + 2hxy + by² + 2gx + 2fy + c = 0.

→ Directrix of the Conic: A line L = 0 passing through the focus of a conic is said to be the principal axis of the conic if it is perpendicular to the directrix of the conic.

→ Vertices: The points of intersection of a conic and its principal axis are called vertices of the conic.

→ Centre: The midpoint o the line segment joining the vertices of a conic is called centre of the conic.

→ Note 1: If a conic has only one vertex then its centre coincides with the vertex.

→ Note 2: If a conic has two vertices then its centre does not coincide either of the vertices. In this case the conic is called a central conic.

→ Standard Form: A conic is said to be in the standard form if the principal axis of the conic is x-axis and the centre of the conic is the origin.

Equation of a Parabola in Standard Form:
The equation of a parabola in the standard form is y² = 4ax.
Proof :
Let S be the focus and L = 0 be the directrix of the parabola.
Let P be a point on the parabola.
Let M, Z be the projections of P, S on the directrix L = 0 respectively.
Let N be the projection of P on SZ.
Let A be the midpoint of SZ.
Therefore, SA = AZ, ⇒ A lies on the parabola. Let AS = a.
Let AS, the principal axis of the parabola as x-axis and Ay perpendicular to SZ as y-axis.
Then S = (a, 0) and the parabola is in the standard form.
Let P = (x₁, y₁).

Now PM = NZ = NA + AZ = x₁ + a
P lies on the parabola ⇒ \(\frac{\mathrm{PS}}{\mathrm{PM}}\) = 1 ⇒ PS = PM
⇒ \(\sqrt{\left(x_{1}-a\right)^{2}+\left(y_{1}-0\right)^{2}}\) = x₁ + a
⇒ (x₁ – a)² + y₁² = (x₁ + a)²
⇒ y₁² = (x₁ + a)² – (x₁ – a)²
⇒ y₁² = 4ax₁
The locus of P is y² = 4ax.
∴ The equation to the parabola is y² = 4ax.

Nature of the Curve y² = 4ax.

(i) The curve is symmetric with respect to the x-axis.
∴ The principal axis (x-axis) is an axis of the parabola.

(ii) y = 0 ⇒ x = 0. Thus the curve meets x-axis at only one point (0, 0).
Hence the parabola has only one vertex.

(iii) If x < 0 then there exists no y ∈ R. Thus the parabola does not lie in the second and third quadrants. (iv) If x > 0 then y² > 0 and hence y has two real values (positive and negative). Thus the parabola lies in the first and fourth quadrants.

(v) x = 0 ⇒ y² = 0 ⇒ y = 0, 0. Thus y-axis meets the parabola in two coincident points and hence y-axis touches the parabola at (0, 0).

(vi) As x → ∞ ⇒ y² → ∞ ⇒ y → ±∞
Thus the curve is not bounded (closed) on the right side of the y-axis.

→ Double Ordinate: A chord passing through a point P on the parabola and perpendicular to the principal axis of the parabola is called the double ordinate of the point P.

→ Focal Chord: A chord of the parabola passing through the focus is called a focal chord.

→ Latus Rectum: A focal chord of a parabola perpendicular to the principal axis of the parabola is called latus rectum. If the latus rectum meets the parabola in L and L’, then LL’ is called length of the latus rectum.

Theorem: The length of the latus rectum of the parabola y² = 4ax is 4a.
Proof:
Let LL’ be the length of the latus rectum of the parabola y² = 4ax.

Let SL = 1, then L = (a, 1)
Since L is a point on the parabola y² = 4ax, therefore 1² = 4a(a)
⇒ 1² = 4a² ⇒ 1 = 2a ⇒ SL = 2a
∴ LL’ = 2SL = 4a.

→ Focal Distance: If P is a point on the parabola with focus S, then SP is called focal distance of P.

Theorem: The focal distance of P(x₁, y₁) on the parabola y² = 4ax is x₁ + a.
Notation: We use the following notation in this chapter
S ≡ y² – 4ax
S₁ ≡ YY₁ – 2a(x + x₁)
S₁₁ = S(x₁, y₁) ≡ y₁² – 4ax₁
S₁₂ ≡ y₁y₂ – 2a(x₁ + x₂)

Note:
Let P(x₁, y₁) be a point and S ≡ y² – 4ax = 0 be a parabola. Then

• P lies on the parabola ⇔ S₁₁ = 0
• P lies inside the parabola ⇔ S₁₁ = 0
• P lies outside the parabola ⇔ S₁₁ = 0

Theorem: The equation of the chord joining the two points A(x₁, y₁), B(x₂, y₂) on the parabola S = 0 is S₁ + S₂ = S₁₂.
Theorem: The equation of the tangent to the parabola S = 0 at P(x₁, y₁) is S₁ = 0.

Normal:
Let S = 0 be a parabola and P be a point on the parabola S = 0. The line passing through P and perpendicular to the tangent of S = 0 at P is called the normal to the parabola S = 0 at P.

Theorem: The equation of the normal to the parabola y² = 4ax at P(x₁, y₁) is y₁(x – x₁) + 2a(y – y₁) = 0.
Proof:
The equation of the tangent to S = 0 at P is S₁ = 0

⇒ yy₁ – 2a(x + x₁) = 0.
⇒ yy₁ – 2ax – 2ax₁ = 0
The equation of the normal to S = 0 at P is:
y₁(x – x₁) + 2a(y – y₁) = 0

Theorem: The condition that the line y = mx + c may be a tangent to the parabola y² = 4ax is c = a/m.
Proof:
Equation of the parabola is y² = 4ax ……….. (1)
Equation of the line is y = mx + c ………. (2)
Solving (1) and (2),
(mx + c)² = 4ax ⇒ m²x² + c² + 2mcx = 4ax
⇒ m²x² + 2(mc – 2a)x + c² = 0
Which is a quadratic equation in x. Therefore it has two roots.
If (2) is a tangent to the parabola, then the roots of the above equation are equal.
⇒ its disc eminent is zero
⇒ 4(mc – 2a)² – 4m²c² = 0
⇒ m²c² + 4a² – 4amc – m²c² = 0
⇒ a² – amc = 0
⇒ a = mc
⇒ C = \(\frac{a}{m}\)

II Method:
Given parabola is y² = 4ax.
Equation of the tangent is y = mx + c ———— (1)
Let P(x₁, y₁) be the point of contact.
The equation of the tangent at P is
yy₁ – 2a(x + x₁) = 0 ⇒ yy₁ = 2ax + 2ax₁ ……. (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{y}_{1}}{1}=\frac{2 \mathrm{a}}{\mathrm{m}}=\frac{2 \mathrm{ax}_{1}}{\mathrm{c}}\) ⇒ x₁ = \(\frac{\mathrm{c}}{\mathrm{m}}\), y₁ = \(\frac{\mathrm{2a}}{\mathrm{m}}\)
P lies on the line y = mx + c ⇒ y₁ = mx₁ + c
⇒ \(\frac{2 \mathrm{a}}{\mathrm{m}}=\mathrm{m}\left(\frac{\mathrm{c}}{\mathrm{m}}\right)+\mathrm{c} \Rightarrow \frac{2 \mathrm{a}}{\mathrm{m}}=2 \mathrm{c} \Rightarrow \mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)

Note: The equation of a tangent to the parabola y² = 4ax can be taken as y = mx + a/m. And the point of contact is (a/m², 2a/m).

Corollary: The condition that the line lx + my + n = 0 to touch the parabola y² = 4ax is am² = ln.
Proof:
Equation of the parabola is y² = 4ax …………. (1)
Equation of the line is lx + my + n = 0
⇒ y = – \(\frac{1}{\mathrm{~m}}\)x – \(\frac{\mathrm{n}}{\mathrm{m}}\)
But this line is a tangent to the parabola, therefore
C = a/m ⇒ \(-\frac{\mathrm{n}}{\mathrm{m}}=\frac{\mathrm{a}}{-1 / \mathrm{m}} \Rightarrow \frac{\mathrm{n}}{\mathrm{m}}=\frac{\mathrm{am}}{1}\) ⇒ am² = ln
Hence the condition that the line lx + my + n = 0 to touché the parabola y² = 4ax is am² = ln.

Note: The point of contact of lx + my + n = 0 with y² = 4ax is (n/l, – 2am/l).

Theorem: The condition that the line lx + my + n = 0 to touch the parabola x² = 4ax is al² = mn.
Proof:
Given line is lx + my + n = 0 …… (1)
Let P(x₁, y₁) be the point of contact of (1) with the parabola x² = 4ay.
The equation of the tangent at P to the parabola is xx₁ = 2a(y + y₁)
⇒ x₁x – 2ay – 2ay₁ = 0 …… (2)
Now (1) and (2) represent the same line.
∴ \(\frac{\mathrm{x}_{1}}{1}=-\frac{2 \mathrm{a}}{\mathrm{m}}=-\frac{2 \mathrm{ay}_{1}}{\mathrm{n}}\) ⇒ x₁ = – \(-\frac{2 \mathrm{al}}{\mathrm{m}}\), y₁ = \(\frac{n}{m}\)
P lies on the line lx + my + n = 0
⇒ lx₁ + my₁ + n = 0 ⇒ l\(\left(\frac{-2 a l}{m}\right)\) + m\(\left(\frac{\mathrm{n}}{\mathrm{m}}\right)\) + n = 0
⇒ – 2al² + mn + mn = 0 ⇒ al² = mn

Theorem: Two tangents can be drawn to a parabola from an external point.

Note:
1. If m₁, m₂ are the slopes of the tangents through P, then m₁, m₂ become the roots of
equation (1). Hence m₁ + m₂ = y₁/x₁, m₁m₂ = a/x₁.

2. If P is a point on the parabola S =0 then the roots of equation (1) coincide and hence only one tangent can be drawn to the parabola through P.

3. If P is an internal point to the parabola S = 0 then the roots of (1) are imaginary and hence no tangent can be drawn to the parabola through P.

Theorem: The equation in the chord of contact of P(x₁, y₁) with respect to the parabola S = 0 is S₁ = 0.

Theorem: The equation of the chord of the parabola S = 0 having P(x₁, y₁) as its midpoint is S₁ = S₁₁.

Pair of Tangents:
Theorem: The equation to the pair of tangents to the parabola S = 0 from P(x₁, y₁) is S₁² = s₁₁s.

Parametric Equations of the Parabola:
A point (x, y) on the parabola y² = 4ax can be represented as x = at², y = 2at in a single parameter t. Theses equations are called parametric equations of the parabola y² = 4ax. The point (at², 2at) is simply denoted by t.

Theorem: The equation of the tangent at (at², 2at) to the parabola is y² = 4ax is yt = x + at².
Proof:
Equation of the parabola is y² = 4ax.
Equation of the tangent at (at², 2at) is S₁ = 0.
⇒ (2at)y – 2a(x + at²) = 0
⇒ 2aty = 2a(x + at²) ⇒ yt = x + at².

Theorem: The equation of the normal to the parabola y² = 4ax at the point t is y + xt = 2at + at³.
Proof:
Equation of the parabola is y² = 4ax.
The equation of the tangent at t is:
yt = x + at² = x – yt + at² = 0
The equation of the normal at (at², 2at) is
t(x – at²) + l(y – 2at) = 0
⇒ xt – at³ + y – 2at = 0 ⇒ y + xt = 2at + at³

Theorem: The equation of the chord joining the points t₁ and t₂ on the parabola y² = 4ax is y(t₁ + t₂) = 2x + 2at₁t₂.
Proof:
Equation of the parabola is y² = 4ax.
Given points on the parabola are
P(at₁², 2at₁), Q(at₂², 2at₂) .
Slope of \(\overline{\mathrm{PQ}}\) is
\(\frac{2 \mathrm{at}_{2}-2 \mathrm{at}_{1}}{\mathrm{at}_{2}^{2}-\mathrm{at}_{1}^{2}}=\frac{2 \mathrm{a}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)}{\mathrm{a}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right)}=\frac{2}{\mathrm{t}_{1}+\mathrm{t}_{2}}\)
The equation of \(\overline{\mathrm{PQ}}\) is y – 2at₁ = \(\frac{2}{t_{1}+t_{2}}\) (x – at₁²).
⇒ (y – 2at₁) (t₁ + t₂) = 2(x – at₁²)
⇒ y(t₁ + t₂) – 2at₁² – 2at₁t₂ = 2x – 2at₁²
⇒ y(t₁ + t₂) = 2x + 2at₁t₂.

Note: If the chord joining the points t₁ and t₂ on the parabola y² = 4ax is a focal chord then t₁t₂ = – 1.
Proof:
Equation of the parabola is y² = 4ax
Focus S = (a, o)
The equation of the chord is y(t₁ + t₂) = 2x + 2at₁t₂
If this is a focal chord then it passes through the focus (a, 0).
∴ 0 = 2a + 2at₁t₂ ⇒ t₁t₂ = – 1.

Theorem: The point of intersection of the tangents to the parabola y² = 4ax at the
points t₁ and t₂ is (at₁t₂, a[t₂ + t₂]).
Proof:
Equation of the parabola is y² = 4ax
The equation of the tangent at t₁ is yt₁ = x + at₁² ……. (1)
The equation of the tangent at t₂ is
yt₂ = x + at₂² ……….. (2)
(1) – (2) ⇒ y(t₁ – t₂) = a(t₁² – t₂²) ⇒ y = a(t₁ + t₂)
(1) ⇒ a(t₁ + t₂)t₁ = x + at₁²
= at₁² + at₁t₁ = x + at₁² ⇒ x = at₁t₂,
∴ Point of intersection = (at₁t₂, a[t₁ + t₂]).

Theorem: Three normals can be drawn form a point (x₁, y₁) to the parabola y² = 4ax.
Corollary: If the normal at t₁ and t₂, to the parabola y² = 4ax meet on the parabola, then t₁t₂ = 2.
Proof:
Let the normals at t₁ and t₂ meet at t₃ on the parabola.
The equation of the normal at t1 is:
y + xt₁ = 2at₁ + at₁³ ………… (1)
Equation of the chord joining t1 and t3 is:
y(t₁ + t₃) = 2x + 2at₁t₃ ……… (2)

(1) and (2) represent the same line
∴ \(\frac{t_{1}+t_{3}}{1}=\frac{-2}{t_{1}} \Rightarrow t_{3}=-t_{1}-\frac{2}{t_{1}}\)
Similarly t₃ = – t₁ – \(\frac{2}{\mathrm{t}_{2}}\)
∴ \(-\mathrm{t}_{1}-\frac{2}{\mathrm{t}_{1}}=-\mathrm{t}_{2}-\frac{2}{\mathrm{t}_{2}}\) ⇒ t₁ – t₂ \(\frac{2}{\mathrm{t}_{2}}-\frac{2}{\mathrm{t}_{1}}\)
⇒ t₁ – t₂ = \(\frac{2\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}{\mathrm{t}_{1} \mathrm{t}_{2}}\) ⇒ t₁t₂ = 2

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 1 Circle to solve questions creatively.

Intermediate 2nd Year Maths 2B Circle Formulas

→ The locus of a point in a plane such that its distance from a fixed point in the plane is always the same is called a circle.

→ The equation of a circle with centre (h, k) and radius r is (x – h)² + (y – k)² = r²

→ The equation of a circle in standard form is x² + y² = r².

→ The equation of a circle in general form is x² + y² + 2gx + 2fy + c = 0 and its centre is (-g, -f), radius is \(\sqrt{g^{2}+f^{2}-c}\).

→ The intercept made by x² + y² + 2gx + 2fy + c = 0

• on X-axis is 2\(\sqrt{g^{2}-c}\) if g² > c.
• on Y-axis is 2\(\sqrt{f^{2}-c}\) if f² > c.

→ If the extremities of a diameter of a circle are (x₁, y₁) and (x₂, y₂) then its equation is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0

→ The equation of a circle passing through three non-collinear points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\) = (x² + y²) + \(\left|\begin{array}{lll}
c_{1} & y_{1} & 1 \\
c_{2} & y_{2} & 1 \\
c_{3} & y_{3} & 1
\end{array}\right|\) x + \(\left|\begin{array}{lll}
x_{1} & C_{1} & 1 \\
x_{2} & C_{2} & 1 \\
x_{3} & C_{3} & 1
\end{array}\right|\) y + \(\left|\begin{array}{lll}
x_{1} & y_{1} & C_{1} \\
x_{2} & y_{2} & C_{2} \\
x_{3} & y_{3} & C_{3}
\end{array}\right|\) = 0.
where c_i = – (x_i² + y_i²)

→ The centre of the circle passing through three non-collinear points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\left[\frac{\left|\begin{array}{lll}
c_{1} & y_{1} & 1 \\
c_{2} & y_{2} & 1 \\
c_{3} & y_{3} & 1
\end{array}\right|}{(-2)\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|}, \frac{\left|\begin{array}{lll}
x_{1} & c_{1} & 1 \\
x_{2} & c_{2} & 1 \\
x_{3} & c_{3} & 1
\end{array}\right|}{(-2)\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|}\right]\)

→ The parametric equations of a circle with centre (h, k) and radius (r ≥ 0) are given by
x = h + r cos θ
y = k + r sin θ 0 ≤ 6 < 2π.

→ A point P(x₁, y₁) is an interior point or on the circumference or an exterior point of a circles S = 0 ⇔ S₁₁ \(\frac{<}{>}\) 0.

→ The power of P(x₁, y₁) with respect to the circle S = 0 is S₁₁.

→ A point P(x₁, y₁) is an interior point or on the circumference or exterior point of the circle S = 0 ⇔ the power of P with respect to S = 0 is negative, zero and positive.

→ If a straight line through a point P(x₁, y₁) meets the circle S = 0 at A and B then the power of P is equal to PA. PB.

→ The length of the tangent from P(x₁, y₁) to S = 0 is \(\sqrt{S_{11}}\).

→ The straight line l = 0 intersects, touches or does not meet the circles = 0 according as l < r, l = r or l > r where l is the perpendicular distance from the centre of the circle to the line l = 0 and r is the radius.

→ For every real value of m the straight line y = mx ± r \(\sqrt{1+m^{2}}\) is a tangent to the circle x² + y² = r².

→ If r is the radius of the circle S = x² + y² + 2gx + 2fy + c = 0 then for every real value of m the straight line y + f = m(n + g) ±r + m2 will be a tangent to the circle.

→ If P(x₁, y₁) and Q(x₂, y₂) are two points on the circle S = 0 then the secant’s \((\stackrel{\leftrightarrow}{P Q})\) equation is S₁ + S₂ = S₁₂

→ The equation of tangent at (x₁, y₁) of the circle S = 0 is S₁ = 0.

→ If θ₁, θ₂ are two points on S = x² + y² + 2gx + 2fy + c = 0 then the equation of the chord joining the points θ₁, θ₂ is
(x + g) cos \(\left(\frac{\theta_{1}+\theta_{2}}{2}\right)\) + (y + f) sin \(\left(\frac{\theta_{1}+\theta_{2}}{2}\right)\) = r cos \(\left(\frac{\theta_{1}-\theta_{2}}{2}\right)\)

→ The equation of the tangent at θ of the circle S = 0 is (x + g) cos θ + (y + f) sin θ = r.

→ The equation of normal at (x₁, y₁) of the circle
S = 0 is (x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0.

→ The chord of contact of P(x₁ y₁) (exterior point) with respect to S = 0 is S₁ = 0.

→ The equation of the polar of a point P(x₁, y₁) with respect to S = 0 is S₁ = 0.

→ The pole of lx + my + n = 0 with respect to S = 0 is
\(\left(-g+\frac{l r^{2}}{l g+m f-n},-f+\frac{m r^{2}}{l g+m f-n}\right)\)

→ Where r is the radius of the circle. The polar of P(x₁, y₁) with respect to S = 0 passes through Q(x₂, y₂) ⇔ the polar of Q with respect to S – 0 passes through P.

→ The points (x₁, y₁) and (x₂, y₂) are conjugate points with respect to S = 0 if S₁₂ = 0

→ Two lines l₁x = m₁y + n₁ = 0, l₂x + m₂y + n₂ = 0 are conjugate with respect to x² + y² = a² ⇔ (l₁l₂ + m₁m₂) = n₁n₂

→ Two points P, Q are said to be inverse points with respect to S = 0 if CP. CQ = r² where C is the centre and r is the radius of the circle S = 0.

→ If (x₁, y₁) is the mid-point of a chord of the circle S = 0 then its chord equation is S₁ = S₁₁.

→ The pair of common tangents to the circles S = 0, S’ = 0 touching at a point on the lines segment \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) (C₁, C₂ are centres of the circles) is called transverse pair of common tangents.

→ The pair of common tangents to the circles S = 0, S’ = 0 intersecting at a point not in \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) is called as direct pair of common tangents.

→ The point of intersection of transvese (direct) common tangents is called internal (external) Centre of similitude.

→ The combined equation of the pair of tangents drawn from an external point P(x₁, y₁) to the circle S = 0 is SS₁₁ = S²₁.

Equation of a Circle:
The equation of the circle with centre C (h, k) and radius r is (x – h)² + (y – k)² = r².
Proof:
Let P(x₁, y₁) be a point on the circle.
P lies in the circle ⇔ PC = r ⇔ \(\sqrt{\left(\mathrm{x}_{1}-\mathrm{h}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{k}\right)^{2}}\) = r
⇔ (x₁ – h)² + (y₁ – k)² = r².

The locus of P is (x – h)² + (y – k)² = r².
∴ The equation of the circle is (x – h)² + (y – k)² = r².

Note: The equation of a circle with centre origin and radius r is (x – 0)² + (y – 0)² = r²
i.e., x² + y² = r² which is the standard equation of the circle.

Note: On expanding equation (1), the equation of a circle is of the form x² + y² + 2gx + 2fy + c = 0.

Theorem: If g² + f² – c ≥ 0, then the equation x² + y² + 2gx + 2fy + c = 0 represents a circle with centre (- g, – f) and radius \(\sqrt{g^{2}+f^{2}-c}\).
Note: If ax² + ay² + 2gx + 2fy + c = 0 represents a circle, then its centre = \(\left(-\frac{g}{a},-\frac{f}{a}\right)\) and its radius \(\frac{\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{ac}}}{|\mathrm{a}|}\).

Theorem: The equation of a circle having the line segment joining A(x₁, y₁) and B(x₂, y₂) as diameter is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0.

Let P(x, y) be any point on the circle. Given points A(x₁, y₁) and B(x₂, y₂).
Now ∠APB = \(\frac{\pi}{2}\). (Angle in a semi circle.)
Slope of AP. Slope of BP = – 1
⇒ \(\frac{y-y_{1}}{x-x_{1}} \frac{y-y_{2}}{x-x_{2}}\) = – 1
⇒ (y – y₂) (y – y₁) = – (x – x₂) (x – x₁) = 0
⇒ (x – x₂) (x – x₁) + (y – y₂) (y – y₁) = 0

Definition: Two circles are said to be concentric if they have same center.

The equation of the circle concentric with the circle x² + y² + 2gx + 2fy + c = 0 is of the form x² + y² + 2gx + 2fy + k = 0.
The equation of the concentric circles differs by constant only.

Parametric Equations of A Circle:

Theorem: If P(x, y) is a point on the circle with centre C(α, β) and radius r, then x = α + r cosθ, y = β + r sin θ where 0 ≤ θ < 2π.

Note: The equations x = α + r cos θ, y = + r sin θ, 0 ≤ θ < 2π are called parametric equations of the circle with centre (α, β) and radius r.

Note: A point on the circle x² + y² = r² is taken in the form (r cos θ, r sin θ). The point (r cos θ, r sin θ) is simply denoted as point θ.

Theorem:
(1) If g² – c > 0 then the intercept made on the x axis by the circle x² + y² + 2gx + 2fy + c = 0 is 2\(\sqrt{g^{2}-a c}\)
(2) If f² – c >0 then the intercept made on the y axis by the circle x² + y² + 2gx + 2fy + c = 0 is 2\(\sqrt{f^{2}-b c}\)

Note: The condition for the x-axis to touch the circle
x² + y² + 2gx + 2fy + c = 0 (c > 0) is g² = c.

Note: The condition of the y-axis to touch the circle
x² + y² + 2gx + 2fy + c = 0 (c > 0) is f² = c.

Position of Point:
Let S = 0 be a circle and P(x₁, y₁) be a point I in the plane of the circle. Then

• P lies inside the circle S = 0 ⇔ S₁₁ < 0
• P lies in the circle S = 0 ⇔ S₁₁ = 0
• Plies outside the circle S = 0 ⇔ S₁₁ = 0

Power of a Point:
Let S = 0 be a circle with centre C and radius r. Let P be a point. Then CP² – r² is called power of P with respect to the circle S = 0.

Theorem: The power of a point P(x₁, y₁) with respect to the circle S = 0 is S₁₁.

Theorem: The length of the tangent drawn from an external point P(x₁, y₁) to the circle s = 0 is \(\sqrt{\mathrm{S}_{11}}\).

Theorem: The equation of the tangent to the circle S = 0 at P(x₁, y₁) is S₁ = 0.

Theorem: The equation of the normal to the circle S = x² + y² + 2gx + 2fy + c = 0 at P(x₁, y₁) is
(y₁ + f) (x – x₁) – (x₁ + g) (y – y₁) = 0.