Q50 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
For some $\theta \in\left(0, \frac{\pi}{2}\right)$, let the eccentricity and the length of the latus rectum of the hyperbola $x^{2}-y^{2} \sec ^{2} \theta=8$ be $e_{1}$ and $l_{1}$, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^{2} \sec ^{2} \theta+y^{2}=6$ be $e_{2}$ and $l_{2}$, respectively. If $e_{1}^{2}=e_{2}^{2}\left(\sec ^{2} \theta+1\right)$, then $\left(\frac{l_{1} l_{2}}{e_{1} e_{2}}\right) \tan ^{2} \theta$ is equal to $\_\_\_\_$
Correct answer: 8
Solution
$\dfrac{x^2}{8} - \dfrac{y^2}{8\cos^2\theta} = 1$, $e_1 = \sqrt{1 + \dfrac{8\cos^2\theta}{8}}$
$\ell_1 = \dfrac{2b^2}{a} = \dfrac{2 \cdot (8\cos^2\theta)}{2\sqrt{2}}$
$\dfrac{x^2}{6} + \dfrac{y^2}{6\cos^2\theta} = 1$; $e_2 = \sqrt{1 - \dfrac{6\cos^2\theta}{6}} = \sin\theta$
$\ell_2 = \dfrac{2b^2}{a} = \dfrac{2 \cdot 6\cos^2\theta}{\sqrt{6}}$
$e_1^2 = e_2^2(1 + \sec^2\theta)$
$1 + \cos^2\theta = \sin^2\theta\left(1 + \dfrac{1}{\cos^2\theta}\right)$
$1 + \cos^2\theta = \sin^2\theta + \tan^2\theta$
Solving we get $\theta = \dfrac{\pi}{4}$
$\ell_1 = 2\sqrt{2}$
$e_1 = \sqrt{\dfrac{3}{2}}$
$\ell_2 = \sqrt{6}$
$e_2 = \dfrac{1}{\sqrt{2}}$
$\left(\dfrac{\ell_1 \ell_2}{e_1 e_2}\right) \tan^2\theta = 8$ (By putting values)
$\ell_1 = \dfrac{2b^2}{a} = \dfrac{2 \cdot (8\cos^2\theta)}{2\sqrt{2}}$
$\dfrac{x^2}{6} + \dfrac{y^2}{6\cos^2\theta} = 1$; $e_2 = \sqrt{1 - \dfrac{6\cos^2\theta}{6}} = \sin\theta$
$\ell_2 = \dfrac{2b^2}{a} = \dfrac{2 \cdot 6\cos^2\theta}{\sqrt{6}}$
$e_1^2 = e_2^2(1 + \sec^2\theta)$
$1 + \cos^2\theta = \sin^2\theta\left(1 + \dfrac{1}{\cos^2\theta}\right)$
$1 + \cos^2\theta = \sin^2\theta + \tan^2\theta$
Solving we get $\theta = \dfrac{\pi}{4}$
$\ell_1 = 2\sqrt{2}$
$e_1 = \sqrt{\dfrac{3}{2}}$
$\ell_2 = \sqrt{6}$
$e_2 = \dfrac{1}{\sqrt{2}}$
$\left(\dfrac{\ell_1 \ell_2}{e_1 e_2}\right) \tan^2\theta = 8$ (By putting values)
Hence, the correct answer is 8.