Question

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 1/2. If the length of the latus rectum of the ellipse E , then the value of 113 l is equal to _____.

Answer 1

The
Vertices of hyperbola = (0, ± 8)
As ellipse pass through it i.e.,
$0+\frac{64}{b^2} = 1$
$⇒ b^2 = 64...(1)$
As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.
e_E = \sqrt{1 - \frac{a^2}{64}}$$= \frac{\sqrt{64-a^2}}{8}
and $e_H = \sqrt{1+\frac{49}{64}} = \frac{\sqrt{113}}{8}$
$∴ e_E . e_H = \frac{1}{2}\frac{\sqrt{64-a^2}\sqrt{113}}{64}$
$⇒ (64-a^2)(113) = 32^2$
$⇒ a^2 = 64-\frac{1024}{113}$
L.R of ellipse $= \frac{2a^2}{b}$
$= \frac{2}{8}(\frac{113×64-1024}{113})$
$I = \frac{1552}{113}$
$∴ 113l = 1552$