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Hyperbola

Section: Mathematics

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Question : 45 of 98

Marks: +1, -0

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

, respectively. Let the product of the eccentricities of

E

H

\;\frac{1}{2}

l

is the length of the latus rectum of the ellipse

E

, then the value of

113 l

is equal to ________.

[27-Jul-2022-Shift-1]

Your Answer:

Solution:

Vertices of hyperbola

=(0, \pm 8)

As ellipse pass through it i.e.,

0+\frac{64}{b^2}=1 \Rightarrow b^2=64 . . . . . . \;\text{(1)}\;

As major axis of ellipse coincide with transverse axis of hyperbola we have

b>a

i.e.

e_E=\sqrt{1-\;\frac{a^2}{64}}=\;\frac{\sqrt{64-a^2}}{8}

and

e_H=\sqrt{1+\;\frac{49}{64}}=\;\frac{\sqrt{113}}{8}

\therefore e_E \cdot e_H=\;\frac{1}{2}=\;\frac{\sqrt{64-a^2} \sqrt{113}}{64}

\Rightarrow (64-a^2)(113)=32^2

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

L.R of ellipse

=\;\frac{2 a^2}{b}=\;\frac{2}{8}\left(\;\frac{113 \times 64-1024}{113}\right)

=l=\;\frac{1552}{113}

\therefore 113 l=1552

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