Question

Let $e_{1}$ be eccentricity of ellipse

$$\Rightarrow e_{1} = \sqrt{1 - \frac{16}{36}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3} $$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. $5 = 2\sqrt{5}$

$$p = \frac{2}{\sqrt{5}} \Rightarrow e^{2} = 1 + \frac{q^{2}}{p^{2}} \Rightarrow 25 = 1 + \frac{5q^{2}}{4} \Rightarrow q^{2} = \frac{96}{5} $$So length of $LR\frac{2q^{2}}{p} = \frac{96}{\sqrt{5}}$