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 113l is equal to _____.

Answer 1

Correct Answer: 1552

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 \(= \frac{2a^2}{b}\)
\(= \frac{2}{8}(\frac{113×64-1024}{113})\)
\(I = \frac{1552}{113}\)
\(∴ 113l = 1552\)