Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 5 Hyperbola to help strengthen their preparations for exams.

## TS Inter 2nd Year Maths 2B Hyperbola Important Questions

Short Answer Type Questions

Question 1.

If e, e_{1} are the eccentricities of a hyperbola and its conjugate hyperbola, prove that \(\frac{1}{e^2}+\frac{1}{e_1^2}=1\)

Solution:

Let e, e_{1} be the eccentricities of hyperbola

Question 2.

Find the equations of the tangents to the hyperbola 3x^{2} – 4y^{2} = 12 which are (I) parallel and (H) perpendicular to the line y = x – 7.

Solution:

Equation of given hyperbola is \(\frac{x^2}{4}-\frac{y^2}{3}=1\)

So that a^{2} = 4, b^{2} = 3 and equation to the given line y = x – 7 and slope is ‘1’.

(i) Slope of the tangents which are parallel to the given line is ‘1’.

∴ Equation of tangents are

y = mx ± \(\sqrt{a^2 m^2-b^2}\)

⇒ y=x± \(\sqrt{4-3}\) and

⇒ y = x ± 1

(ii) Slope of the tangent which are perpendicular to the given line is – 1.

∴ Equations of tangents which are perpendicular to the given line are

y = (-1) x ± \(\sqrt{4(-1)^2-3}\)

x+y = ±1

Question 3.

A circle on the rectangular hyperbola xy = 1. In the points (Xr Yr)’ (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 \((\mathrm{t}, \frac{1}{t})\)(t ≠ 0) lies on xy= 1, the points of intersection of the circle and the hyperbola are given by

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

Long Answer Type Questions

Question 1.

Find the centre, eccentricity, foci, directrices and the length of latus rectum of the following hyperbolas.

4x^{2} – 9y^{2} – 8x -32 = 0

Solution:

Given equation is 4x^{2 }– 9y^{2} – 8x – 32 = 0

⇒ 4x^{2} – 8x -9y^{2} = 32

⇒ 4(x -2x)-9y=32

⇒ 4(x^{2}-2x+ 1) – 9y^{2 }= 32 + 4 = 36

⇒ 4 (x-1)2 – 9y^{2} = 36

\(\frac{(x-1)^2}{9}-\frac{(y-0)^2}{4}=1\)

∴ Centre of the hyperbola = (1, 0)

The semi-transverse axis a = 3, and the semiconjugate axis b = 2.

∴ \(e=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{9+4}{9}}=\sqrt{\frac{13}{3}}\)

Coordinates of foci (h ± ae, k)

(ii) 4(y+3)^{2 }-9(x-2)^{2}= 1

Solution:

The equation can be written as

Question 2.

(i) If t be line lx+ my+n= 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 = t is a normal to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) then show that

\(\frac{a^2}{l^2}-\frac{b^2}{m^2}=\left(a^2+b^2\right)^2\)

Solution:

(i) Let the line lx + my + n = 0 ……………….. (1) is a tangent to the hyperbola S = \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) at P(θ).

Then the equation of tangent at P(θ) is

\(\frac{x}{a}\) secθ – \(\frac{x}{b}\) tanθ – 1 = θ ………….. (2)

Since (1) and (2) represent the same line,

Question 3.

Prove that the point of intersection of two per perpendicular tangents to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) lies on the circle x +y =a – b.

Solution:

Let P (x_{1}, y_{1}) be a point of intersection of two perpendicular tangents to the hyperbola

The equation of any tangent to S = 0 is of the form

\(S \equiv \frac{x^2}{a^2}-\frac{y^2}{b^2}-1=0\)

The equation of any tangent to S = 0 is of the form

This is a quadratic equation ¡n ‘m’ which has two roots m_{1}, m_{2} (say) which corresponds to slopes of tangents.

Question 4.

If your 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 tanα, tanβ, tanγ , tanδ = 1

Solution:

Let the equation of rectangular hyperbola be x^{2} – y^{2} = a^{2}. By rotating the X – axis and Y – axis about the orgin through an angle \(\frac{\pi}{4}\) in the clockwise direction the equation x^{2} – y^{2} = a^{2} will be transformed to xy = C^{2}.

Since \(\overline{\mathrm{AB}}\) is perpendicular \(\overline{\mathrm{CD}}\) we have

\(\left(-\frac{1}{t_1 t_2}\right)\left(-\frac{1}{t_3 t_4}\right)=-1\)

⇒ t_{1} t_{1 }t_{1 }t_{1 }= – 1 ………………. (1)

We have the coordinate axis as the a asymptotes of the curves.

If \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}, \overline{\mathrm{OC}}, \overline{\mathrm{OD}}\) make angles α, β, γ and δ with positive direction of X-axis then tanα, tanβ, tanγ, and tanδ are the slopes.

If \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}, \overline{\mathrm{OC}}, \overline{\mathrm{OD}}\) make angles α, β, γ and δ with the other asymptote the Y – axis then cot α, cot β, cot γ, cot δ are the respective slopes.

So that cot α cot β cot γ cot δ = tan α tan β tan γ tan δ = I