JEE Main 2026 — 21 January, Morning Shift — Question 11

Let $\mathrm{e}_{1}$ be eccentricity of ellipse $\Rightarrow e_{1}=\sqrt{1-\frac{16}{36}}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$

So $\mathrm{ae}_{1}=6 \cdot \frac{\sqrt{5}}{3}=2 \sqrt{5}$

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

$pe=\mathrm{ae}_{1}$

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

$\mathrm{p}=\frac{2}{\sqrt{5}}$ $\Rightarrow \mathrm{e}^{2}=1+\frac{\mathrm{q}^{2}}{\mathrm{p}^{2}}$

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

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

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