JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 18)
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 $$E$$ and $$H$$ be $$\frac{1}{2}$$. If $$l$$ is the length of the latus rectum of the ellipse $$E$$, then the value of $$113 l$$ is equal to _____________.
Answer
1552
Explanation
Vertices of hyperbola $$ = (0,\, \pm \,8)$$
As ellipse pass through it i.e.,
$$0 + {{64} \over {{b^2}}} = 1 \Rightarrow {b^2} = 64$$ ...... (1)
As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.
$${e_E} = \sqrt {1 - {{{a^2}} \over {64}}} = {{\sqrt {64 - {a^2}} } \over 8}$$
and $${e_H} = \sqrt {1 + {{49} \over {64}}} = {{\sqrt {113} } \over 8}$$
$$\therefore$$ $${e_E}\,.\,{e_H} = {1 \over 2} = {{\sqrt {64 - {a^2}} \sqrt {113} } \over {64}}$$
$$ \Rightarrow (64 - {a^2})(113) = {32^2}$$
$$ \Rightarrow {a^2} = 64 - {{1024} \over {113}}$$
L.R of ellipse $$ = {{2{a^2}} \over b} = {2 \over 8}\left( {{{113 \times 64 - 1024} \over {113}}} \right)$$
$$ = l = {{1552} \over {113}}$$
$$\therefore$$ $$113l = 1552$$
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