Let e1 and e2 be the eccentricities of the ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$ and the hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$, respectively. If b < 5 and e1e2 = 1, then the ecc

Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci. Step 1: Find $ e_1 $ for the Ellipse The equation of the ellipse is: $ \frac{x^2}{b^2} + \frac{y^2}{25} = 1 $ The eccentricity $ e_1 $ is given by: $ e_1^2 = 1 - \frac{b^2}{25} $ Step 2: Find $ e_2 $ for the Hyperbola The equation of the hyperbola is: $ \frac{x^2}{16} - \frac{y^2}{b^2} = 1 $ The eccentricity $ e_2 $ is given by: $ e_2^2 = 1 + \fra

Hyperbola — Page 5 | JEE Mathematics

Q41

Let

P

be a point on the hyperbola

H: \dfrac{x^2}{9}-\dfrac{y^2}{4}=1

, in the first quadrant such that the area of triangle formed by

P

and the two foci of

H

is

2 \sqrt{13}

. Then, the square of the distance of

P

from the origin is

A 26

B 22

C 20

D 18

Correct Answer

Option B

Solution

\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{4}=1 \\ & a^2=9, b^2=4 \\ & b^2=a^2\left(e^2-1\right) \Rightarrow e^2=1+\frac{b^2}{a^2} \\ & e^2=1+\frac{4}{9}=\frac{13}{9} \\ & e=\frac{\sqrt{13}}{3} \Rightarrow s_1 s_2=2 \mathrm{ae}=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13} \end{aligned}

\begin{aligned} & \text{ Area of } \triangle \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{2} \times \beta \times \mathrm{S}_1 \mathrm{~S}_2=2 \sqrt{13} \\ & \Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 \\ & \begin{aligned} & \frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2} \\ & \text{ Distance of P from origin }=\sqrt{\alpha^2+\beta^2} \\ &=\sqrt{18+4}=\sqrt{22} \end{aligned} \end{aligned}

Q42

Let the foci of a hyperbola

H

coincide with the foci of the ellipse

E: \dfrac{(x-1)^2}{100}+\dfrac{(y-1)^2}{75}=1

and the eccentricity of the hyperbola

H

be the reciprocal of the eccentricity of the ellipse

E

. If the length of the transverse axis of

H

is

\alpha

and the length of its conjugate axis is

\beta

, then

3 \alpha^2+2 \beta^2

is equal to

A 225

B 237

C 242

D 205

Correct Answer

Option A

Solution

E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1

\begin{aligned} \text{ Eccentricity of ellipse, } e_E & =\sqrt{1-\frac{b^2}{a^2}} \\ & =\sqrt{1-\frac{75}{100}} \\ & e_E=\frac{1}{2} \end{aligned}

\therefore e_H=2

[ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse] Transverse axis of hyperbola

=\alpha

Conjugate axis of hyperbola

=\beta

Also, foci of ellipse

(1 \pm a e, 1)

\begin{aligned} & =\left(1 \pm\left(10 \times \frac{1}{2}\right), 1\right) \\ & =(1 \pm 5,1) \\ & =(6,1) \text{ and }(-4,1) \end{aligned}

Distance between foci

=10

\begin{aligned} & 2 a e=10 \\ & \Rightarrow a=\frac{5}{2} \end{aligned}

\begin{aligned} & \text{ also, } e^2=1+\frac{b^2}{a^2} \\ & \begin{aligned} 4 & =1+\frac{4 b^2}{25} \\ b^2 & =\frac{75}{4} \\ b & =\frac{\sqrt{75}}{2} \end{aligned} \\ & \Rightarrow \quad \alpha=5 \\ & \text{ and } \beta=\sqrt{75} \\ & 3 \alpha^2+2 \beta^2=3(5)^2+2(75)=225 \end{aligned}

Q43

Consider a hyperbola

\mathrm{H}

having centre at the origin and foci on the

\mathrm{x}

-axis. Let

\mathrm{C}_1

be the circle touching the hyperbola

\mathrm{H}

and having the centre at the origin. Let

\mathrm{C}_2

be the circle touching the hyperbola

\mathrm{H}

at its vertex and having the centre at one of its foci. If areas (in sq units) of

C_1

and

C_2

are

36 \pi

and

4 \pi

, respectively, then the length (in units) of latus rectum of

\mathrm{H}

is

A

\dfrac{28}{3}

B

\dfrac{11}{3}

C

\dfrac{14}{3}

D

\dfrac{10}{3}

Correct Answer

Option A

Solution

\begin{aligned} & C_1: x^2+y^2=a^2 \Rightarrow \text{ area }=\pi a^2=36 \pi \Rightarrow a=6 \\ & C_2: x^2+(y-a e)^2=(a e-a)^2 \\ & \therefore \quad \pi(a e-a)^2=4 \pi \\ & \Rightarrow 36(e-1)^2=4 \\ & \Rightarrow e-1=\frac{1}{3} \Rightarrow e=\frac{4}{3} \\ & \Rightarrow b^2=28 \\ & \therefore \quad L R=\frac{2 b^2}{a}=\frac{2 \times 28}{6}=\frac{28}{3} \text{ units } \end{aligned}

Q44

Let

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

be the hyperbola, whose eccentricity is

\sqrt{3}

and the length of the latus rectum is

4 \sqrt{3}

. Suppose the point

(\alpha, 6), \alpha>0

lies on

H

. If

\beta

is the product of the focal distances of the point

(\alpha, 6)

, then

\alpha^2+\beta

is equal to

A 170

B 171

C 169

D 172

Correct Answer

Option B

Solution

\begin{aligned} & H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\ & e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3} \\ & \Rightarrow 1+\frac{a^2}{b^2}=3 \\ & \Rightarrow \frac{a^2}{b^2}=2 \quad \text{.... (1)}\\ & \frac{2 a^2}{b}=4 \sqrt{3} \end{aligned}

Using equation (1)

\begin{aligned} & \frac{4 b^2}{b}=4 \sqrt{3} \\ & \Rightarrow b=\sqrt{3} \\ & a=\sqrt{6} \\ & H: \frac{x^2}{6}-\frac{y^2}{3}=-1 \\ & \frac{\alpha^2}{6}-12=-1 \\ & \frac{\alpha^2}{6}=11 \\ & \begin{array}{c} \alpha^2=66 \\ \text{ Focus }:(0, b c)(0,-b c) \\ (0,3),(0,-3) \end{array} \\ & \beta=\sqrt{\alpha^2+9} \times \sqrt{\alpha^2+81} \\ & \beta=105 \\ & \alpha^2+\beta=66+105 \\ & =171 \end{aligned}

Q45

Let the foci of a hyperbola be

(1,14)

and

(1,-12)

. If it passes through the point

(1,6)

, then the length of its latus-rectum is :

A

\dfrac{25}{6}

B

\dfrac{144}{5}

C

\dfrac{288}{5}

D

\dfrac{24}{5}

Correct Answer

Option C

Solution

$\begin{aligned} & \mathrm{be}=13, \mathrm{~b}=5 \\ & \mathrm{a}^2=\mathrm{b}^2\left(\mathrm{e}^2-1\right) \\ & =\mathrm{b}^2 \mathrm{e}^2-\mathrm{b}^2 \\ & =169-25=144 \\ & \ell(\mathrm{LR})=\dfrac{2 \mathrm{a}^2}{\mathrm{~b}}=\dfrac{2 \times 144}{5}=\dfrac{288}{5}\end{aligned}$

Q46

Let one focus of the hyperbola

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

be at

(\sqrt{10}, 0)

and the corresponding directrix be

x=\dfrac{9}{\sqrt{10}}

. If

e

and

l

respectively are the eccentricity and the length of the latus rectum of H , then

9\left(e^2+l\right)

is equal to :

A 12

B 14

C 15

D 16

Correct Answer

Option D

Solution

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

Directrix : $x=\dfrac{9}{\sqrt{10}}=\dfrac{a}{e}\quad\text{..... (i)}$ Focus : $(\sqrt{10}, 0) \equiv(a e, 0)$

a e=\sqrt{10}\quad\text{..... (ii)}

(i) $\times$ (ii)

\Rightarrow a^2=9 \Rightarrow a=3

Substitute in (ii)

e=\frac{\sqrt{10}}{3}

\begin{aligned} & \text{ Now } e^2=1+\frac{b^2}{a^2} \\ & \frac{10}{9}=1+\frac{b^2}{a} \\ & \Rightarrow b=1 \\ & I=\frac{2 b^2}{a}=\frac{2 \times 1}{3}=\frac{2}{3} \\ & a\left[e^2+l\right]=9\left[\frac{10}{9}+\frac{2}{3}\right]=10+6 \\ & =16 \end{aligned}

Q47

Let e1 and e2 be the eccentricities of the ellipse

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

and the hyperbola

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

, respectively. If b < 5 and e1e2 = 1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

A

\dfrac{4}{5}

B

\dfrac{3}{5}

C

\dfrac{\sqrt{7}}{4}

D

\dfrac{\sqrt{3}}{2}

Correct Answer

Option B

Solution

Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci.

Step 1: Find $e_1$ for the Ellipse The equation of the ellipse is: $\dfrac{x^2}{b^2} + \dfrac{y^2}{25} = 1$ The eccentricity $e_1$ is given by: $e_1^2 = 1 - \dfrac{b^2}{25}$ Step 2: Find $e_2$ for the Hyperbola The equation of the hyperbola is: $\dfrac{x^2}{16} - \dfrac{y^2}{b^2} = 1$ The eccentricity $e_2$ is given by: $e_2^2 = 1 + \dfrac{b^2}{16}$ Step 3: Using the Product $e_1 e_2 = 1$ Given: $e_1 e_2 = 1$ Thus: $\left(1 - \dfrac{b^2}{25}\right)\left(1 + \dfrac{b^2}{16}\right) = 1$ Expanding gives: $1 + \dfrac{b^2}{16} - \dfrac{b^2}{25} - \dfrac{b^4}{400} = 1$ Simplifying: $\dfrac{9b^2}{400} = \dfrac{b^4}{400}$ Thus: $b^2 = 9$ Step 4: Determine Eccentricities $e_1$ and $e_2$ Substitute $b^2 = 9$ : For the ellipse: $e_1^2 = 1 - \dfrac{9}{25} = \dfrac{16}{25}$ $e_1 = \dfrac{4}{5}$ For the hyperbola: $e_2 = \dfrac{5}{4}$ Step 5: Find the Eccentricity of the New Ellipse The new ellipse's equation is: $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ The eccentricity $e$ is: $e = \sqrt{1 - \dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$ Thus, the eccentricity of the ellipse that passes through all four foci is $\dfrac{3}{5}$ .

Q48

Let the sum of the focal distances of the point

\mathrm{P}(4,3)

on the hyperbola

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

be

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

. If for H , the length of the latus rectum is

l

and the product of the focal distances of the point P is m , then

9 l^2+6 \mathrm{~m}

is equal to :

A 187

B 184

C 186

D 185

Correct Answer

Option D

Solution

\begin{aligned} &\sqrt{(c+4)^2+9}+\sqrt{(c-4)^2+9}=8 \sqrt{\frac{5}{3}}\\ &\text{ Solving, } c=\frac{5}{\sqrt{6}}=\mathrm{al} \Rightarrow a^2\left(1+\frac{b^2}{a^2}\right)=\frac{25}{6} \Rightarrow a^2+b^2=\frac{25}{6}\\ &\begin{aligned} & \frac{16}{a^2}-\frac{9}{b^2}=1 \\ & 16 b^2-9 d^2=a^2 b^2 \Rightarrow 16\left(\frac{25}{6}-a^2\right)-9 a^2=9 a^2 b^2 \\ & P F_1+P F_2=8 \sqrt{\frac{5}{3}} \Rightarrow a^2=\frac{5}{2}, b^2=\frac{5}{3} \\ & \left|P F_1-P F_2\right|=2 a \\ & \frac{64.5}{3}=4 a^2+4 m \Rightarrow m=\frac{80}{3}-a^2 \\ & 6 m=160-6 a^2 \end{aligned} \end{aligned}

\begin{aligned} & 9 \ell^2=9\left(\frac{2 b^2}{a}\right)^2=\frac{36 b^4}{a^2} \\ & 9 \ell^2+6 m=\frac{36\left(\frac{25}{9}\right)}{\frac{5}{2}}+160-6\left(\frac{5}{2}\right) \\ & =\frac{72 \times 5}{9}+160-15 \\ & =160+40-15=185 \end{aligned}

Q49

Let the tangent drawn to the parabola

y^{2}=24 x

at the point

(\alpha, \beta)

is perpendicular to the line

2 x+2 y=5

. Then the normal to the hyperbola

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

at the point

(\alpha+4, \beta+4)

does NOT pass through the point :

A (25, 10)

B (20, 12)

C (30, 8)

D (15, 13)

Correct Answer

Option D

Solution

Any tangent to

{y^2} = 24x

at ( $\alpha$ , $\beta$ )

\beta y = 12(x + \alpha )

Slope

= {{12} \over \beta }

and perpendicular to

2x + 2y = 5

\Rightarrow {{12} \over \beta } = 1 \Rightarrow \beta = 12,\,\alpha = 6

Hence hyperbola is

{{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1

and normal is drawn at (10, 16) Equation of normal

{{36\,.\,x} \over {10}} + {{144\,.\,y} \over {16}} = 36 + 144

\Rightarrow {x \over {50}} + {y \over {20}} = 1

This does not pass though (15, 13) out of given option.

Q50

Let a and b respectively be the semitransverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9e2 − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a2 − b2 is equal to :

A 7

B

-

7

C 5

D

-

5

Correct Answer

Option B

Solution

As S(5, 0) is the focus. $\therefore$ ae = 5 . . . (1) As 5x = 9 $\therefore$ x =

{9 \over 5}

is the directrix As we know directrix =

{a \over e}

$\therefore$

{a \over e} = {9 \over 5}

. . . .(2) Solving (1) and (2), we get a = 3 and e =

{5 \over 3}

As we know, b2 = a2 (e2 $-$ 1) = 9

\left( {{{25} \over 9} - 1} \right)

= 16 $\therefore$ a2 $-$ b2 = 9 $-$ 16 $=$ $-$ 7