An ellipse E: x ^ 2 a ^ 2 + y ^ 2 b ^ 2 = 1 passes through the...

Given,

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

And ellipse $E: \dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ passes through the vertices of the hyperbola $H: \dfrac{{x}^{2}}{49} - \dfrac{{y}^{2}}{64} = - 1$, so vertices will be $V ≡ \left(0, \pm 8\right)$

So ${b}^{2} = 64$

Now eccentricity of hyperbola will be ${e}_{H} = \sqrt{1 + \dfrac{{a}^{2}}{{b}^{2}}} = \sqrt{1 + \dfrac{49}{64}}$

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

${e}_{E} = \sqrt{1 - \dfrac{{a}^{2}}{{b}^{2}}} = \sqrt{1 - \dfrac{{a}^{2}}{64}}$

And using $b = 8$

We get, ${e}_{H} \times {e}_{E} = \dfrac{1}{2}$ (given)

$⇒ \sqrt{\dfrac{1 - {a}^{2}}{64}} \times \dfrac{\sqrt{113}}{8} = \dfrac{1}{2}$

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

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

$⇒ {a}^{2} = 64 - \dfrac{{32}^{2}}{113}$

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

$⇒ 113l = 1552$