Hyperbola — Page 4 | JEE Mathematics

Q31

The normal to the hyperbola

{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1

at the point

\left( {8,3\sqrt 3 } \right)

on it passes through the point :

A

\left( {15, - 2\sqrt 3 } \right)

B

\left( {9,2\sqrt 3 } \right)

C

\left( { - 1,9\sqrt 3 } \right)

D

\left( { - 1,6\sqrt 3 } \right)

Correct Answer

Option C

Solution

Given hyperbola :

{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1

$\because$ It passes through

(8,3\sqrt 3 )

$\because$

{{64} \over {{a^2}}} - {{27} \over 9} = 1 \Rightarrow {a^2} = 16

Now, equation of normal to hyperbola

{{16x} \over 8} + {{9y} \over {3\sqrt 3 }} = 16 + 9

\Rightarrow 2x + \sqrt 3 y = 25

...... (i)

\left( { - 1,9\sqrt 3 } \right)

satisfies (i)

Q32

Let the foci of the ellipse

\dfrac{x^{2}}{16}+\dfrac{y^{2}}{7}=1

and the hyperbola

\dfrac{x^{2}}{144}-\dfrac{y^{2}}{\alpha}=\dfrac{1}{25}

coincide. Then the length of the latus rectum of the hyperbola is :

A

\dfrac{32}{9}

B

\dfrac{18}{5}

C

\dfrac{27}{4}

D

\dfrac{27}{10}

Correct Answer

Option D

Solution

Ellipse :

{{{x^2}} \over {16}} + {{{y^2}} \over 7} = 1

Eccentricity

= \sqrt {1 - {7 \over {16}}} = {3 \over 4}

Foci

\equiv ( \pm \,a\,e,0) \equiv ( \pm \,3,0)

Hyperbola :

{{{x^2}} \over {\left( {{{144} \over {25}}} \right)}} - {{{y^2}} \over {\left( {{\alpha \over {25}}} \right)}} = 1

Eccentricity

= \sqrt {1 + {\alpha \over {144}}} = {1 \over {12}}\sqrt {144 + \alpha }

Foci

\equiv ( \pm \,a\,e,0) \equiv \left( { \pm \,{{12} \over 5}\,.\,{1 \over {12}}\sqrt {144 + \alpha } ,\,0} \right)

If foci coincide then

3 = {1 \over 5}\sqrt {144 + \alpha } \Rightarrow \alpha = 81

Hence, hyperbola is

{{{x^2}} \over {{{\left( {{{12} \over 5}} \right)}^2}}} - {{{y^2}} \over {{{\left( {{9 \over 5}} \right)}^2}}} = 1

Length of latus rectum

= 2\,.\,{{{{81} \over {25}}} \over {{{12} \over 5}}} = {{27} \over {10}}

Q33

If the line

x-1=0

is a directrix of the hyperbola

k x^{2}-y^{2}=6

, then the hyperbola passes through the point :

A

(-2 \sqrt{5}, 6)

B

(-\sqrt{5}, 3)

C

(\sqrt{5},-2)

D

(2 \sqrt{5}, 3 \sqrt{6})

Correct Answer

Option C

Solution

Given hyperbola :

{{{x^2}} \over {6/k}} - {{{y^2}} \over 6} = 1

Eccentricity

= e = \sqrt {1 + {6 \over {6/k}}} = \sqrt {1 + k}

Directrices :

x = \, \pm \,{a \over e} \Rightarrow x = \, \pm \,{{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }}

As given :

{{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }} = 1

\Rightarrow k = 2

Here hyperbola is

{{{x^2}} \over 3} - {{{y^2}} \over 6} = 1

Checking the option gives

\left( {\sqrt 5 , - 2} \right)

satisfies it.

Q34

Let the hyperbola

H: \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1

pass through the point

(2 \sqrt{2},-2 \sqrt{2})

. A parabola is drawn whose focus is same as the focus of

\mathrm{H}

with positive abscissa and the directrix of the parabola passes through the other focus of

\mathrm{H}

. If the length of the latus rectum of the parabola is e times the length of the latus rectum of

\mathrm{H}

, where e is the eccentricity of H, then which of the following points lies on the parabola?

A

(2 \sqrt{3}, 3 \sqrt{2})

B

\mathbf(3 \sqrt{3},-6 \sqrt{2})

C

(\sqrt{3},-\sqrt{6})

D

(3 \sqrt{6}, 6 \sqrt{2})

Correct Answer

Option B

Solution

H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

Focus of parabola :

(ae,\,0)

Directrix :

x = - ae

. Equation of parabola

\equiv {y^2} = 4aex

Length of latus rectum of parabola

= 4ae

Length of latus rectum of hyperbola

= {{2.{b^2}} \over a}

as given,

4ae = {{2{b^2}} \over a}\,.\,e

2 = {{{b^2}} \over {{a^2}}}

...... (i) $\because$ H passes through

\left( {2\sqrt 2 , - 2\sqrt 2 } \right) \Rightarrow {8 \over {{a^2}}} - {8 \over {{b^2}}} = 1

........ (ii) From (i) and (ii)

{a^2} = 4

and

{b^2} = 8 \Rightarrow e = \sqrt 3

$\Rightarrow$ Equation of parabola is

{y^2} = 8\sqrt 3 x

.

Q35

Let

\mathrm{P}\left(x_{0}, y_{0}\right)

be the point on the hyperbola

3 x^{2}-4 y^{2}=36

, which is nearest to the line

3 x+2 y=1

. Then

\sqrt{2}\left(y_{0}-x_{0}\right)

is equal to :

A 3

B

-

9

C

-

3

D 9

Correct Answer

Option B

Solution

If $\left(x_0, y_0\right)$ is point on hyperbola then tangent at $\left(x_0, y_0\right)$ is parallel to $3 x+2 y=1$ Equation of tangent $= \dfrac{x x_0}{12}-\dfrac{y y_0}{9}=2$ Slope of tangent $=\dfrac{-3}{2}$ Equation of tangent in slope form $y=\dfrac{-3}{2} x \pm \sqrt{12 \cdot \dfrac{9}{4}-9}$ $y=\dfrac{-3}{2} x \pm 3 \sqrt{2}$ or $3 x+2 y=6 \sqrt{2}$ Comparing

\begin{aligned} & \frac{\frac{x_0}{12}}{3}=\frac{\frac{-y_0}{9}}{2}=\frac{1}{6 \sqrt{2}} \\\\ & x_0=3 \sqrt{2}, y_0=\frac{-3}{\sqrt{2}} \\\\ & \sqrt{2}\left(y_0-x_0\right)=-3-6=-9 \\\\ & \end{aligned}

Q36

Let T and C respectively be the transverse and conjugate axes of the hyperbola

16{x^2} - {y^2} + 64x + 4y + 44 = 0

. Then the area of the region above the parabola

{x^2} = y + 4

, below the transverse axis T and on the right of the conjugate axis C is :

A

4\sqrt 6 - {{28} \over 3}

B

4\sqrt 6 - {{44} \over 3}

C

4\sqrt 6 + {{28} \over 3}

D

4\sqrt 6 + {{44} \over 3}

Correct Answer

Option C

Solution

$16(x+2)^{2}-(y-2)^{2}=16$

\frac{(x+2)^{2}}{1}-\frac{(y-2)^{2}}{16}=1

TA $: y=2$ $\mathrm{CA}: x=-2$

A=\left|\int_{-2}^{\sqrt{6}}\left(2-\left(x^{2}-4\right)\right) d x\right|

\begin{aligned} & =6 x-\left.\frac{x^{3}}{3}\right|_{-2} ^{\sqrt{6}} \\\\ & =\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right) \end{aligned}

=\frac{12 \sqrt{6}}{3}+\frac{28}{3}

Q37

Let R be a rectangle given by the lines

x=0, x=2, y=0

and

y=5

. Let A

(\alpha,0)

and B

(0,\beta),\alpha\in[0,2]

and

\beta\in[0,5]

, be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the mid-point of AB lies on a :

A hyperbola

B straight line

C parabola

D circle

Correct Answer

Option A

Solution

We have, $R$ be a rectangle formed by the lines $x=0$ , $x=2, y=0$ and $y=5$ Let $P Q R S$ be the rectangle such that

P(0,0), Q(2,0), R(2,5) \text{ and } S(0,5)

$P Q=S R=2$ units and $P S=Q R=5$ units $\therefore$ Area of rectangle $P Q R S=(2 \times 5) \mathrm{sq}$ units $=10 \mathrm{sq}$ units Area of $\triangle P A B=\dfrac{1}{5} \times$ Area of rectangle

\begin{aligned} \frac{1}{2} \times \alpha \times \beta & =\frac{1}{5} \times 10 \text{ sq units } \\\\ & =2 \text{ sq units } \\\\ \frac{1}{2} \times \alpha \times \beta & =2 \\\\ \Rightarrow \alpha \beta & =4 \text{ sq units } .........(i) \end{aligned}

Let $M(h, k)$ be the mid-point of $A B$ .

Here, $h=\dfrac{\alpha}{2}, k=\dfrac{\beta}{2}$ or $\alpha=2 h$ and $\beta=2 k$ On substitute values of $\alpha$ and $\beta$ in Eq. (i), we get

\begin{aligned} \quad(2 h)(2 k) & =4 \\\\ \Rightarrow h k & =1 \\\\ \therefore \text{ Locus of } M \text{ is } x y & =1 \text{, which is a hyperbola. } \end{aligned}

Q38

For

0<\theta<\pi / 2

, if the eccentricity of the hyperbola

x^2-y^2 \operatorname{cosec}^2 \theta=5

is

\sqrt{7}

times eccentricity of the ellipse

x^2 \operatorname{cosec}^2 \theta+y^2=5

, then the value of

\theta

is :

A

\dfrac{\pi}{6}

B

\dfrac{5 \pi}{12}

C

\dfrac{\pi}{3}

D

\dfrac{\pi}{4}

Correct Answer

Option C

Solution

To find the value of $\theta$ , we need to determine the relationship between the eccentricities of the given hyperbola and ellipse.

Let's start by writing down the standard forms of ellipse and hyperbola and then relate them to the given equations.

The standard form of an ellipse is: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ The eccentricity $e$ of an ellipse is given by: $e = \sqrt{1 - \dfrac{b^2}{a^2}}\ \text{for}\ a > b$ For the given ellipse: $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$ We can compare it with the standard form by writing it as: $\dfrac{x^2}{\dfrac{5}{\operatorname{cosec}^2 \theta}} + \dfrac{y^2}{5} = 1$ From this we have $a^2 = \dfrac{5}{\operatorname{cosec}^2 \theta}$ and $b^2 = 5$ .

The eccentricity of the ellipse (let's call it $e_1$ ) is: $e_1 = \sqrt{1 - \dfrac{5}{a^2}} = \sqrt{1 - \dfrac{5}{\dfrac{5}{\operatorname{cosec}^2 \theta}}} = \sqrt{1 - \operatorname{sin}^2 \theta} = \operatorname{cos} \theta$ The standard form of a hyperbola is: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ The eccentricity $e$ of a hyperbola is given by: $e = \sqrt{1 + \dfrac{b^2}{a^2}}$ For the given hyperbola: $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$ We can rewrite it as: $\dfrac{x^2}{5} - \dfrac{y^2}{\dfrac{5}{\operatorname{cosec}^2 \theta}} = 1$ From this we have $a^2 = 5$ and $b^2 = \dfrac{5}{\operatorname{cosec}^2 \theta}$ .

The eccentricity of the hyperbola (let's call it $e_2$ ) is: $e_2 = \sqrt{1 + \dfrac{\dfrac{5}{\operatorname{cosec}^2 \theta}}{5}} = \sqrt{1 + \operatorname{sin}^2 \theta}$ According to the question the eccentricity of the hyperbola is $\sqrt{7}$ times the eccentricity of the ellipse, so we can write: $e_2 = \sqrt{7} \cdot e_1$ Substitute $e_1$ and $e_2$ from the above: $\sqrt{1 + \operatorname{sin}^2 \theta} = \sqrt{7} \cdot \operatorname{cos} \theta$ Square both sides to eliminate the square root: $1 + \operatorname{sin}^2 \theta = 7 \cdot \operatorname{cos}^2 \theta$ $1 + \operatorname{sin}^2 \theta = 7 \cdot (1 - \operatorname{sin}^2 \theta)$ $1 + \operatorname{sin}^2 \theta = 7 - 7 \cdot \operatorname{sin}^2 \theta$ $8 \cdot \operatorname{sin}^2 \theta = 6$ $\operatorname{sin}^2 \theta = \dfrac{6}{8} = \dfrac{3}{4}$ $\operatorname{sin} \theta = \sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}$ Looking at the options provided and knowing the sine values for common angles, we can see that $\operatorname{sin} \theta = \dfrac{\sqrt{3}}{2}$ corresponds to $\theta = \dfrac{\pi}{3}$ .

The correct answer is: Option C.

$\dfrac{\pi}{3}$

Q39

Let

e_1

be the eccentricity of the hyperbola

\dfrac{x^2}{16}-\dfrac{y^2}{9}=1

and

e_2

be the eccentricity of the ellipse

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, \mathrm{a} > \mathrm{b}

, which passes through the foci of the hyperbola. If

\mathrm{e}_1 \mathrm{e}_2=1

, then the length of the chord of the ellipse parallel to the

x

-axis and passing through

(0,2)

is :

A

\dfrac{8 \sqrt{5}}{3}

B

3 \sqrt{5}

C

4 \sqrt{5}

D

\dfrac{10 \sqrt{5}}{3}

Correct Answer

Option D

Solution

\begin{aligned} & H: \frac{x^2}{16}-\frac{y^2}{9}=1 \qquad e_1=\frac{5}{4} \\ & \therefore e_1 e_2=1 \Rightarrow e_2=\frac{4}{5} \end{aligned}

Also, ellipse is passing through

( \pm 5,0)

\begin{aligned} & \therefore a=5 \text{ and } b=3 \\ & E: \frac{x^2}{25}+\frac{y^2}{9}=1 \end{aligned}

End point of chord are

\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)

\therefore \mathrm{L}_{\mathrm{PQ}}=\frac{10 \sqrt{5}}{3}

Q40

If the foci of a hyperbola are same as that of the ellipse

\dfrac{x^2}{9}+\dfrac{y^2}{25}=1

and the eccentricity of the hyperbola is

\dfrac{15}{8}

times the eccentricity of the ellipse, then the smaller focal distance of the point

\left(\sqrt{2}, \dfrac{14}{3} \sqrt{\dfrac{2}{5}}\right)

on the hyperbola, is equal to

A

14 \sqrt{\dfrac{2}{5}}-\dfrac{4}{3}

B

7 \sqrt{\dfrac{2}{5}}+\dfrac{8}{3}

C

7 \sqrt{\dfrac{2}{5}}-\dfrac{8}{3}

D

14 \sqrt{\dfrac{2}{5}}-\dfrac{16}{3}

Correct Answer

Option C

Solution

\begin{aligned} & \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \\ & \mathrm{a}=3, \mathrm{~b}=5 \\ & \mathrm{e}=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text{ foci }=(0, \pm \mathrm{be})=(0, \pm 4) \\ & \quad \therefore \mathrm{e}_{\mathrm{H}}=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2} \end{aligned}

Let equation hyperbola

\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1 \\ & \therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3} \\ & \therefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^2=\frac{80}{9} \\ & \therefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1 \\ & \text{ Directrix }: \mathrm{y}= \pm \frac{\mathrm{B}}{\mathrm{e}_{\mathrm{H}}}= \pm \frac{16}{9} \\ & \mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right| \\ & =7 \sqrt{\frac{2}{5}}-\frac{8}{3} \end{aligned}