An ellipse E : x 2 a 2 + y 2 b 2 = 1 passes through the vertices of the hyperbola H : x 2 49 - y 2 64 = - 1 …

Given,

Hyperbola:$\frac{y^2}{64}-\frac{x^2}{49}=1$

And 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 vertices will be $V\equiv \left(0,\pm 8\right)$

So $b^2=64$

Now eccentricity of hyperbola will be $e_H=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+\frac{49}{64}}$

And eccentricity of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ will be

$e_E=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{a^2}{64}}$

And using $b=8$

We get, $e_H\times e_E=\frac{1}{2}$ (given)

$\Rightarrow \sqrt{\frac{1-a^2}{64}}\times \frac{\sqrt{113}}{8}=\frac{1}{2}$

$\Rightarrow \sqrt{64-a^2}\times \sqrt{113}=32$

$\Rightarrow \left(64-a^2\right)=\frac{{32}^2}{113}$

$\Rightarrow a^2=64-\frac{{32}^2}{113}$

Now length of latus rectum will be $l=\frac{2a^2}{b}=\frac{2}{8}\left(64-\frac{{32}^2}{113}\right)=\frac{1552}{113}$

$\Rightarrow 113l=1552$