JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Ellipse

An ellipse

E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

passes through the vertices of the hyperbola

H : \frac{x ^{2}}{49} - \frac{y ^{2}}{64} = - 1

. Let the major and minor axes of the ellipse E coincide with the transverse and conjugate axes of the hyperbola H. Let the product of the eccentricities of

E

and

H

\frac{1}{2}

. If

l

is the length of the latus rectum of the ellipse E, then the value of 113l is equal to ____.

Answer: 1552

Solution

The hyperbola

H : \frac{y ^{2}}{64} - \frac{x ^{2}}{49} = 1

has vertices at

(0, \pm 8)

.

Since the ellipse passes through these vertices and its major axis is along the

y

-axis:

b^{2} = 64

, so

b = 8

.

Eccentricity of hyperbola:

e_{H} = 1 + \frac{49}{64} = \frac{113}{8}

Eccentricity of ellipse:

e_{E} = 1 - \frac{a ^{2}}{64}

Given:

e_{E} \times e_{H} = \frac{1}{2}

1 - \frac{a ^{2}}{64} \times \frac{113}{8} = \frac{1}{2}

64 - a^{2} \times 113 = 32

(64 - a^{2}) \times 113 = 1024

a^{2} = 64 - \frac{1024}{113}

Length of latus rectum:

l = \frac{2 a ^{2}}{b} = \frac{2}{8} (64 - \frac{1024}{113}) = \frac{1}{4} \times \frac{7232 - 1024}{113} = \frac{1552}{113}

Therefore,

113 l = 1552

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Ellipse

Solution

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