JEE Main 2026 — 28 January, Morning Shift — Question 21

$\frac{x^{2}}{8}-\frac{y^{2}}{8 \cos ^{2} \theta}=1$

$e_{1}=\sqrt{1+\frac{8 \cos ^{2} \theta}{8}}$

$\ell_{1}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{2 \cdot\left(8 \cos ^{2} \theta\right)}{2 \sqrt{2}}$

$\frac{x^{6}}{6}+\frac{y^{2}}{6 \cos ^{2} \theta}=1$

$e_{2}=\sqrt{1-\frac{6 \cos ^{2} \theta}{6}}=\sin \theta$

$\ell_{2}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{2.6 \cos ^{2} \theta}{\sqrt{6}}$

$\mathrm{e}_{1}^{2}=\mathrm{e}_{2}^{2}\left(1+\sec ^{2} \theta\right)$

$1+\cos ^{2} \theta=\sin ^{2} \theta\left(1+\frac{1}{\cos ^{2} \theta}\right)$

$1+\cos ^{2} \theta=\sin ^{2} \theta+\tan ^{2} \theta$

Solving we get $\theta=\frac{\pi}{4}$

$\ell_{1}=2 \sqrt{2}$

$\mathrm{e}_{1}=\sqrt{\frac{3}{2}}$

$\ell_{2}=\sqrt{6}$

$\mathrm{e}_{2}=\frac{1}{\sqrt{2}}$

$\left(\frac{\ell_{1} \ell_{2}}{\mathrm{e}_{1} \mathrm{e}_{2}}\right) \tan ^{2} \theta=8 \quad$ (By putting values)