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. 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 $113l$ is equal to ________

Answer 1

Answer: 1552

Hyp : $\frac{y^2}{64}-\frac{x^2}{49}=1$
An ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the vertices of the hyperbola $H:\frac{x^2}{49}-\frac{y^2}{64}=-1$
So $b^2=64$
$e_H=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+\frac{49}{64}}$
Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$e_E=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{a^2}{64}}$
$b=8,\sqrt{\frac{1-a^2}{64}}\times \frac{\sqrt{113}}{8}=\frac{1}{2}\Rightarrow \sqrt{64-a^2}\times \sqrt{113}=32$
$\left(64-a^2\right)=\frac{32^2}{113}$
$\Rightarrow a^2=64-\frac{32^2}{113}$
$l=\frac{2a^2}{b}=\frac{2}{8}\left(64-\frac{32^2}{113}\right)=\frac{1552}{113}$
$113l=1552$