Inter 2nd Year Maths 2B Hyperbola Important Questions

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

Intermediate 2nd Year Maths 2B Hyperbola Important Questions

Question 1.
Find the equations of the hyperbola whose foci are (±5, 0) the transverse axis is of length 8. [T.S. Mar. 16; May 11]
Solution:
Foci are S(±5,0) ∴ ae = 5
Length of transverse axis = 2a = 8
a = 4
e = \(\frac{5}{4}\)
b² = a²(e² – 1) = 16(\(\frac{25}{16}\) – 1) = 9
Equation of the hyperbola is \(\frac{x^{2}}{16}\) – \(\frac{y^{2}}{9}\) = 1
9x² – 16y² = 144.

Question 2.
If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate-hyperbola. [AP Mar. 16] [TS Mar. 15, 13]
Solution:
If e and e₁, are the eccentricity of a hyper bola and its conjugate hyperbola, then
\(\frac{1}{\mathrm{e}^{2}}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1
Given e = \(\frac{5}{4}\) = \(\frac{16}{25}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1
\(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1 – \(\frac{16}{25}\) = \(\frac{9}{25}\) e_1² = \(\frac{25}{9}\)
e₁ = \(\frac{5}{3}\)

Question 3.
Find the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the x² – 4y² = 4 hyperbola. [A.P. Mar. 16; May 11]
Solution:
Equation of the hyperbola is \(\frac{x^{2}}{4}\)– \(\frac{y^{2}}{1}\) = 1
a² = 4, b² = 1
Centre is c (0, 0)
a²e² = a² + b² = 4 + 1 = 5
ae = \(\sqrt{5}\)
Foci are (±ae, 0) = (±\(\sqrt{5}\), 0)
Eccentricity = \(\frac{\mathrm{ae}}{\mathrm{a}}\) = \(\frac{\sqrt{5}}{2}\)
Equations of directrices are x = ± \(\frac{\mathrm{ae}}{\mathrm{a}}\)
= ± 2 . \(\frac{2}{\sqrt{5}}\)
⇒ \(\sqrt{5}\) x = ± 4
⇒ \(\sqrt{5}\) x ± 4 = 0
Length of the latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2.1}{2}\) = 1

Question 4.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are
i) Parallel and
ii) Perpendicular to the line y = x – 7. [AP Mar. 15]
Solution:
i) Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1
a² = 4, b² = 3.
The tangent is parallel to y = x – 7
m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = x ± \(\sqrt{4-3}\)
y = x ± 1

ii) The tangent is perpendicular to y – x = 7
m – slope of the tangent = (-1)
Equation of the perpendicular tangents are
y = (-1) x ± \(\sqrt{4(-1)^{2}-3}\)
= -x ± 1
x + y = ± 1.

Question 5.
If 3x – 4y – k = 0 is a tangent to x² – 4y² = 5, find value of k. [T.S. Mar. 17]
Solution:
Equation of the hyperbola x² – 4y² = 5
\(\frac{x^{2}}{5}\) – \(\frac{y^{2}}{\left(\frac{5}{4}\right)}\) = 1
a² = 5, b² = \(\frac{5}{4}\)
Equation of the given line is 3x — 4y + k = 0
4y = 3x + k
y = \(\frac{3}{4}\) x + \(\frac{k}{4}\)
m = \(\frac{3}{4}\), c = \(\frac{k}{4}\),
Condition for tangency is c² = a²m² – b²
\(\frac{\mathrm{k}^{2}}{16}\) = 5 . \(\frac{9}{16}\) – \(\frac{5}{4}\)
k² = 45 – 20 = 25
k = ± 5.

Question 6.
Find the equations of the tangents to the hyperbola x² – 4y² = 4 Which are
i) Parallel
ii) Perpendicular to the line x + 2y = 0. [T.S. Mar. 15, Mar. 14, 11; May 06]
Solution:
Equation of the hyperbola is
x² – 4y² = 4
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{1}\) = 1
a² = 4, b² = 1
i) The tangent is parallel to x + 2y = 0
m = –\(\frac{1}{2}\)
c² = a²m² – b² = 4 . \(\frac{1}{4}\) = 1 – 1 = 0
c = 0
Equation of the parallel tangent is
y = mx + c
= –\(\frac{1}{2}\) x
2y = -x
x + 2y = 0.

ii) The tangent is perpendicular to x + 2y = 0
Slope of the tangent = m = \(\frac{-1}{\left(-\frac{1}{2}\right)}\) = 2
c² = a²m² – b² = 4 . 4 – 1 = 153
c = ±\(\sqrt{15}\)
Equation of the perpendicular tangent is
y = 2x ± \(\sqrt{15}\)

Question 7.
Prove that 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. [T.S. Mar. 16]
Solution:
Let P (x₁, y₁) be the point of intersection of two perpendicular tangents to the hyperbola
\(\frac{x^{2}}{a^{2}}\) – \(\frac{x^{2}}{b^{2}}\) = 1
Equation of the tangent can be taken as
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P (x₁, y₁)
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ = a²m² – b²
m²x₁² – a²m² – 2mx₁y₁ + y₁² + b² = 0
m²(x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0
This is a quadratic in m giving the values say m₁, m₁ which are the slopes of the tangents
passing through P
The tangents are perpendicular
⇒ m₁m₂ = – 1
\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = – 1 ⇒ y₁² + b²
x₁² + y₁² = a² – b²
focus of P (x₁, y₁) is x² + y² = a² – b².
This circle is called director circle of the hyperbola.

Question 8.
If e, e₁ are the eccentricities of a hyperbola and its conjugate hyperbola prove that \(\frac{1}{e^{2}}\) + \(\frac{1}{\mathrm{e}_{1}^{2}}\) = 1 [Mar. 11]
Solution:
Equation of the hyperbola is \(\frac{x^{2}}{a^{2}}\) – \(\frac{x^{2}}{b^{2}}\) = 1
∴ b² = a²(e² – 1) ⇒ e² – 1 = \(\frac{b^{2}}{a^{2}}\)
e² = 1 + \(\frac{b^{2}}{a^{2}}\) = \(\frac{a^{2}+b^{2}}{a^{2}}\)
∴ \(\frac{1}{\mathrm{e}^{2}}\) = \(\frac{a^{2}}{a^{2}+b^{2}}\) …………… (1)
Equation of the conjugate hyperbola is

Question 9.
Find the centre eccentricity, foci, directrices and length of the latus rectum of the following hyperbolas.
i) 4x² – 9y² – 8x – 32 = 0
ii) 4 (y + 3)² – 9(x – 2)² = 1.
Solution:
i) 4x² – 9y² – 8x – 32 = 0
4(x² – 2x) – 9y² = 32
4(x² – 2x + 1) – 9y² = 36
\(\frac{(x-1)^{2}}{9}\) – \(\frac{(y)^{2}}{4}\) = 1
Centre of the hyperbola is C (1, 0)
a² = 9, b² = 4 ⇒ a = 3, b = 2
e = \(\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{9+4}{9}}=\frac{\sqrt{13}}{3}\)
Foci are (1±3. \(\frac{\sqrt{13}}{3}\), 0) = (1±\(\sqrt{13}\), 0)
Equations of differences are x = 1 ± \(\frac{3.3}{\sqrt{13}}\)
⇒ x = 1 ± \(\frac{9}{\sqrt{13}}\)
Length of the latus rectum = \(\frac{2 b^{2}}{a}\)
= \(\frac{2.4}{3}\) = \(\frac{8}{3}\)

ii) The equation of the hyperbola is
4 (y + 3)² – 9 (x – 2)² = 1
\(\frac{y-(-3)^{2}}{1 / 4}\) = \(\frac{(x-2)^{2}}{1 / 9}\) = 1
Centre is C (2, -3)
Semi transverse axis = b = \(\frac{1}{2}\)
Semi conjugate axis = a = \(\frac{1}{3}\)

Question 10.
If e, e₁ are the eccentricities of a hyperbola and its conjugate hyperbola prove that \(\frac{1}{e^{2}}\) + \(\frac{1}{e_{1}^{2}}\) = 1. [Mar. 11]
Solution:

Equation of the conjugate hyperbola is

Question 11.
i) If the line lx + my = 0 is a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1, then show that a²l² – b²m² = n².

ii) If the lx + my = 1 is a normal to the hyperbola \(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1, then show that \(\frac{a^{2}}{l^{2}}\) – \(\frac{b^{2}}{m^{2}}\) = (a² + b²)².
Solution:
i) Equation of the given tangent ¡s
lx + my + n = 0 ……………. (1)
Equation of the tangent P(θ) is
\(\frac{x}{a}\) sec θ – \(\frac{y}{b}\) tan θ – 1 = 0 …………….. (2)
Comparing (1) and (2)
\(\frac{\sec \theta}{a l}\) = \(\frac{\tan \theta}{-\mathrm{bm}}\) = \(\frac{-1}{n}\)
sec θ = –\(\frac{\mathrm{a} l}{\mathrm{n}}\), tan θ = \(\frac{\mathrm{bm}}{\mathrm{n}}\)
sec² θ – tan²θ = 1
= \(\frac{a^{2} l^{2}}{n^{2}}\) – \(\frac{b^{2} m^{2}}{n^{2}}\) = 1 ⇒ a²l² – b²m² = n².

ii) Equation of the given line is lx + my = 1 ……………. (1)
Equation of the normal at P(θ) is
\(\frac{a x}{\sec \theta}\) + \(\frac{b y}{\tan \theta}\) = a² + b² ……….. (2)
Comparing (1) and (2)

Question 12.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are
i) Parallel and ii) Perpendicular to the line y = x – 7. [A.P. Mar. 15]
Solution:
i) Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1
a² = 4, b² = 3
The tangent is parallel to y = x – 7
m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
y = x ± \(\sqrt{4-3}\)
y = x ± 1

ii) Th tangent is perpendicular to y – x = 7
m – slope of the tangent = (-1)
Equation of the perpendicular tangents are
y = (-1) x ± \(\sqrt{4(-1)^{2}-3}\)
= -x ± 1
x + y = ±1.

Question 13.
Prove that 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. [T.S. Mar. 16]
Solution:
Let P (x₁, y₁) be the point of intersection of two perpendicular tangents to the hyperbola
\(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1
Equation of the tangent can be taken as
y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
This tangent passes through P (x₁, y₁)
y₁ = mx₁ ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(y₁ – mx₁)² = a²m² – b²
y₁² + m²x₁² – 2mx₁y₁ = a²m² – b²
m²x₁² – a²m² – 2mx₁y₁ + y₁² + b² = 0
m² (x₁² – a²) – 2mx₁y₁ + (y₁² + b²) = 0
This is a quadratic in m giving the values say m₁, m₂ which are the slopes of the tangents passing through P
The tangents are perpendicular
⇒ m₁m₂ = – 1
\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = -1 ⇒ y₁² + b² = -x₁² + a²
x₁² + y₁² = a² – b²
focus of P (x₁, y₁) is x² + y² = a² – b².
This circle is called director circle of the hyperbola.

Question 14.
A circle cuts the rectangular hyperbola xy = 1 in the points (x_r, y_r), r = 1, 2, 3, 4. Prove that x₁x₂x₃x₄ = y₁y₂y₃y₄ = 1.
Solution:
Let the circle be x² + y² = a².
Since (t, \(\frac{1}{t}\)) (t ≠ 0) lies on xy = 1, the points of intersection of the circle and the hyperbola are given by
t² + \(\frac{1}{t^{2}}\) = a²
⇒ t⁴ – a²t² + 1 = 0
⇒ t⁴ + 0 . t³ – a²t² + 0 . t + 1 = 0.
If t₁, t₂, t₃ and t₄ are the roots of the above biquadratic, then t₁t₂t₃t₄ = 1.
If (x_r, y_r) = (t_r; \(\frac{1}{t_{r}}\)), r = 1, 2, 3, 4
then x₁x₂x₃x₄ = t₁t₂t₃t₄ = 1,
and y₁y₂y₃y₄ = \(\frac{1}{t_{1} t_{2} t_{3} t_{4}}\) = 1

Question 15.
If four points be taken on a rectangular hyperbola such that the chords joining any two points is perpendicular to the chord joining the other two, and if α, β, γ and δ be the inclinations to either asymptote of the straight lines joining these points to the centre, prove that
tan α tan β tan γ tan δ = 1.
Solution:
Let the equation of the rectangular hyperbola be x² – y² = a². By rotating the X-axis and the Y-axis about the origin through an angle \(\frac{\pi}{4}\) in the clockwise sense, the equation x² – y² = a² can be transformed to the form xy = c².
Let (ct_r, \(\frac{c}{t_{r}}\)), r = 1, 2, 3, 4 (t_r ≠ 0) be four point on the curve. Let the chord joining
A = (ct₁, \(\frac{c}{t_{1}}\)), B = (ct₂, \(\frac{c}{t_{2}}\)) be perpendicular to the chord joining C = (ct₃, \(\frac{c}{t_{3}}\)) and D = (ct₄, \(\frac{c}{t_{4}}\)).
The slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\frac{\frac{c}{t_{1}}-\frac{c}{t_{2}}}{c t_{1}-c t_{2}}=\frac{-1}{t_{1} t_{2}}\)
[No chord of the hyperbola can be vertical]
Similarly slope of \(\stackrel{\leftrightarrow}{C D}\) is –\(\frac{1}{t_{3} t_{4}}\), Since \(\stackrel{\leftrightarrow}{A B}\) ⊥ \(\stackrel{\leftrightarrow}{C D}\).
\(\left(-\frac{1}{t_{1} t_{2}}\right)\left(-\frac{1}{t_{3} t_{4}}\right)\) = -1 ⇒ t₁t₂t₃t₄ = -1 ………………… (1)
We know the coordinate axes are the asymptotes of the curve, If \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) make angles α, β, γ and δ with the positive direction of the X-axis, then tan α, tan β, tan γ and tan δ are their respective slopes. [O, the origin is the centre, None of \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) is vertical]

If \(\stackrel{\leftrightarrow}{\mathrm{OA}}\), \(\stackrel{\leftrightarrow}{\mathrm{OB}}\), \(\stackrel{\leftrightarrow}{\mathrm{OC}}\) and \(\stackrel{\leftrightarrow}{\mathrm{OD}}\) make angles α, β, γ and δ with the other asymptote the Y-axis then cot α, cot β, cot γ and cot δ are their respective inclinations so that
cot α cot β cot γ cot δ = tan α tan β tan γ tan δ = 1.