Let the foci of hyperbola coincide with the foci of the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$. If the eccentricity of the hyperbola is $5$, then the length of its latus rectum is:

Let $e_1$ be eccentricity of ellipse

$\Rightarrow e_1 = \sqrt{1 - \frac{16}{36}} = \sqrt{\frac{4}{9}} = \frac{\sqrt{5}}{3}$

So $ae_1 = 6 \cdot \frac{\sqrt{5}}{3} = 2\sqrt{5}$

Now $H : \frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$

$p.e = ae_1$

$p \cdot 5 = 2\sqrt{5}$

$p = \frac{2}{\sqrt{5}}$

$e^2 = 1 + \frac{q^2}{p^2}$

$25 = 1 + \frac{q^2}{p^2} \Rightarrow 25 = 1 + \frac{5q^2}{4}$

$q^2 = \frac{96}{5}$

So length of LR $= \frac{2q^2}{p} = \frac{96}{\sqrt{5}}$