For some θ ∈(0, (π)/(2)) , let the eccentricity and the length of...

\dfrac{x^2}{8} - \dfrac{y^2}{8\cos^2\theta} = 1$,$e_1 = \sqrt{1 + \dfrac{8\cos^2\theta}{8}}$<br>$\ell_1 = \dfrac{2b^2}{a} = \dfrac{2 \cdot (8\cos^2\theta)}{2\sqrt{2}}$<br>$\dfrac{x^2}{6} + \dfrac{y^2}{6\cos^2\theta} = 1$;$e_2 = \sqrt{1 - \dfrac{6\cos^2\theta}{6}} = \sin\theta$<br>$\ell_2 = \dfrac{2b^2}{a} = \dfrac{2 \cdot 6\cos^2\theta}{\sqrt{6}}$<br>$e_1^2 = e_2^2(1 + \sec^2\theta)$<br>$1 + \cos^2\theta = \sin^2\theta\left(1 + \dfrac{1}{\cos^2\theta}\right)$<br>$1 + \cos^2\theta = \sin^2\theta + \tan^2\theta$<br>Solving we get$\theta = \dfrac{\pi}{4}$<br>$\ell_1 = 2\sqrt{2}$<br>$e_1 = \sqrt{\dfrac{3}{2}}$<br>$\ell_2 = \sqrt{6}$<br>$e_2 = \dfrac{1}{\sqrt{2}}$<br>\left(\dfrac{\ell_1 \ell_2}{e_1 e_2}\right) \tan^2\theta = 8$ (By putting values)