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 = \(\frac{5}{4}\)

b^{2} = a^{2}(e^{2} – 1) = 16(\(\frac{25}{16}\) – 1) = 9

Equation of the hyperbola is \(\frac{x^{2}}{16}\) – \(\frac{y^{2}}{9}\) = 1

9x^{2} – 16y^{2} = 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 e_{1}, 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} = \(\frac{25}{9}\)

e_{1} = \(\frac{5}{3}\)

Question 3.

Find the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the x^{2} – 4y^{2} = 4 hyperbola. [A.P. Mar. 16; May 11]

Solution:

Equation of the hyperbola is \(\frac{x^{2}}{4}\)– \(\frac{y^{2}}{1}\) = 1

a^{2} = 4, b^{2} = 1

Centre is c (0, 0)

a^{2}e^{2} = a^{2} + b^{2} = 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 3x^{2} – 4y^{2} = 12 which are

i) Parallel and

ii) Perpendicular to the line y = x – 7. [AP Mar. 15]

Solution:

i) Equation of the hyperbola is 3x^{2} – 4y^{2} = 12

\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1

a^{2} = 4, b^{2} = 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 x^{2} – 4y^{2} = 5, find value of k. [T.S. Mar. 17]

Solution:

Equation of the hyperbola x^{2} – 4y^{2} = 5

\(\frac{x^{2}}{5}\) – \(\frac{y^{2}}{\left(\frac{5}{4}\right)}\) = 1

a^{2} = 5, b^{2} = \(\frac{5}{4}\)

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 c^{2} = a^{2}m^{2} – b^{2}

\(\frac{\mathrm{k}^{2}}{16}\) = 5 . \(\frac{9}{16}\) – \(\frac{5}{4}\)

k^{2} = 45 – 20 = 25

k = ± 5.

Question 6.

Find the equations of the tangents to the hyperbola x^{2} – 4y^{2} = 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

x^{2} – 4y^{2} = 4

\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{1}\) = 1

a^{2} = 4, b^{2} = 1

i) The tangent is parallel to x + 2y = 0

m = –\(\frac{1}{2}\)

c^{2} = a^{2}m^{2} – b^{2} = 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

c^{2} = a^{2}m^{2} – b^{2} = 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 x^{2} + y^{2} = a^{2} – b^{2}. [T.S. Mar. 16]

Solution:

Let P (x_{1}, y_{1}) 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 (x_{1}, y_{1})

y_{1} = mx_{1} ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

(y_{1} – mx_{1})^{2} = a^{2}m^{2} – b^{2}

y_{1}^{2} + m^{2}x_{1}^{2} – 2mx_{1}y_{1} = a^{2}m^{2} – b^{2}

m^{2}x_{1}^{2} – a^{2}m^{2} – 2mx_{1}y_{1} + y_{1}^{2} + b^{2} = 0

m^{2}(x_{1}^{2} – a^{2}) – 2mx_{1}y_{1} + (y_{1}^{2} + b^{2}) = 0

This is a quadratic in m giving the values say m_{1}, m_{1} which are the slopes of the tangents

passing through P

The tangents are perpendicular

⇒ m_{1}m_{2} = – 1

\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = – 1 ⇒ y_{1}^{2} + b^{2}

x_{1}^{2} + y_{1}^{2} = a^{2} – b^{2}

focus of P (x_{1}, y_{1}) is x^{2} + y^{2} = a^{2} – b^{2}.

This circle is called director circle of the hyperbola.

Question 8.

If e, e_{1} 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

∴ b^{2} = a^{2}(e^{2} – 1) ⇒ e^{2} – 1 = \(\frac{b^{2}}{a^{2}}\)

e^{2} = 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) 4x^{2} – 9y^{2} – 8x – 32 = 0

ii) 4 (y + 3)^{2} – 9(x – 2)^{2} = 1.

Solution:

i) 4x^{2} – 9y^{2} – 8x – 32 = 0

4(x^{2} – 2x) – 9y^{2} = 32

4(x^{2} – 2x + 1) – 9y^{2} = 36

\(\frac{(x-1)^{2}}{9}\) – \(\frac{(y)^{2}}{4}\) = 1

Centre of the hyperbola is C (1, 0)

a^{2} = 9, b^{2} = 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, e_{1} 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 a^{2}l^{2} – b^{2}m^{2} = n^{2}.

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 \(\frac{a^{2}}{l^{2}}\) – \(\frac{b^{2}}{m^{2}}\) = (a^{2} + b^{2})^{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 = \(\frac{\mathrm{bm}}{\mathrm{n}}\)

sec^{2} θ – tan^{2}θ = 1

= \(\frac{a^{2} l^{2}}{n^{2}}\) – \(\frac{b^{2} m^{2}}{n^{2}}\) = 1 ⇒ a^{2}l^{2} – b^{2}m^{2} = n^{2}.

ii) Equation of the given line is lx + my = 1 ……………. (1)

Equation of the normal at P(θ) is

\(\frac{a x}{\sec \theta}\) + \(\frac{b y}{\tan \theta}\) = a^{2} + b^{2} ……….. (2)

Comparing (1) and (2)

Question 12.

Find the equations of the tangents to the hyperbola 3x^{2} – 4y^{2} = 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 3x^{2} – 4y^{2} = 12

\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{3}\) = 1

a^{2} = 4, b^{2} = 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 x^{2} + y^{2} = a^{2} – b^{2}. [T.S. Mar. 16]

Solution:

Let P (x_{1}, y_{1}) 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 (x_{1}, y_{1})

y_{1} = mx_{1} ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

(y_{1} – mx_{1})^{2} = a^{2}m^{2} – b^{2}

y_{1}^{2} + m^{2}x_{1}^{2} – 2mx_{1}y_{1} = a^{2}m^{2} – b^{2}

m^{2}x_{1}^{2} – a^{2}m^{2} – 2mx_{1}y_{1} + y_{1}^{2} + b^{2} = 0

m^{2} (x_{1}^{2} – a^{2}) – 2mx_{1}y_{1} + (y_{1}^{2} + b^{2}) = 0

This is a quadratic in m giving the values say m_{1}, m_{2} which are the slopes of the tangents passing through P

The tangents are perpendicular

⇒ m_{1}m_{2} = – 1

\(\frac{y_{1}^{2}+b^{2}}{x_{1}^{2}-a^{2}}\) = -1 ⇒ y_{1}^{2} + b^{2} = -x_{1}^{2} + a^{2}

x_{1}^{2} + y_{1}^{2} = a^{2} – b^{2}

focus of P (x_{1}, y_{1}) is x^{2} + y^{2} = a^{2} – b^{2}.

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 x_{1}x_{2}x_{3}x_{4} = y_{1}y_{2}y_{3}y_{4} = 1.

Solution:

Let the circle be x^{2} + y^{2} = a^{2}.

Since (t, \(\frac{1}{t}\)) (t ≠ 0) lies on xy = 1, the points of intersection of the circle and the hyperbola are given by

t^{2} + \(\frac{1}{t^{2}}\) = a^{2}

⇒ t^{4} – a^{2}t^{2} + 1 = 0

⇒ t^{4} + 0 . t^{3} – a^{2}t^{2} + 0 . t + 1 = 0.

If t_{1}, t_{2}, t_{3} and t_{4} are the roots of the above biquadratic, then t_{1}t_{2}t_{3}t_{4} = 1.

If (x_{r}, y_{r}) = (t_{r}; \(\frac{1}{t_{r}}\)), r = 1, 2, 3, 4

then x_{1}x_{2}x_{3}x_{4} = t_{1}t_{2}t_{3}t_{4} = 1,

and y_{1}y_{2}y_{3}y_{4} = \(\frac{1}{t_{1} t_{2} t_{3} t_{4}}\) = 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 x^{2} – y^{2} = a^{2}. By rotating the X-axis and the Y-axis about the origin through an angle \(\frac{\pi}{4}\) in the clockwise sense, the equation x^{2} – y^{2} = a^{2} can be transformed to the form xy = c^{2}.

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 = (ct_{1}, \(\frac{c}{t_{1}}\)), B = (ct_{2}, \(\frac{c}{t_{2}}\)) be perpendicular to the chord joining C = (ct_{3}, \(\frac{c}{t_{3}}\)) and D = (ct_{4}, \(\frac{c}{t_{4}}\)).

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 ⇒ t_{1}t_{2}t_{3}t_{4} = -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.