Hyperbola

Concept:We need to find eccentricities and latus recta of a hyperbola and an ellipse, use a given relation to find $\theta$, then compute the required expression.Explanation:Rewrite the hyperbola $x^2 - y^2 \sec^2 \theta = 8$ in standard form.Divide by 8: $\frac{x^2}{8} - \frac{y^2}{8\cos^2\theta} = 1$.Here $a^2 = 8$, $b^2 = 8\cos^2\theta$.Eccentricity of hyperbola: $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \cos^2\theta$.Latus rectum of hyperbola: $l_1 = \frac{2b^2}{a} = \frac{2 \cdot 8\cos^2\theta}{\sqrt{8}} = 4\sqrt{2}\,\cos^2\theta$.Rewrite the ellipse $x^2 \sec^2 \theta + y^2 = 6$ in standard form.Divide by 6: $\frac{x^2}{6\cos^2\theta} + \frac{y^2}{6} = 1$.Since $\theta \in (0,\pi/2)$, $\cos^2\theta < 1$, so $6\cos^2\theta < 6$.Thus the major axis is the $y$-axis: $b^2 = 6$, $a^2 = 6\cos^2\theta$.Eccentricity of ellipse: $e_2^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{6\cos^2\theta}{6} = \sin^2\theta$.Latus rectum of ellipse: $l_2 = \frac{2a^2}{b} = \frac{2 \cdot 6\cos^2\theta}{\sqrt{6}} = 2\sqrt{6}\,\cos^2\theta$.Given condition: $e_1^2 = e_2^2 (\sec^2\theta + 1)$.Substitute: $1 + \cos^2\theta = \sin^2\theta \left( \frac{1}{\cos^2\theta} + 1 \right) = \sin^2\theta \cdot \frac{1 + \cos^2\theta}{\cos^2\theta}$.Cancel $1 + \cos^2\theta$ (nonzero): $1 = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$.So $\tan^2\theta = 1$ and $\theta \in (0,\pi/2)$ gives $\theta = \frac{\pi}{4}$.Now compute $\frac{l_1 l_2}{e_1 e_2} \cdot \tan^2\theta$.At $\theta = \frac{\pi}{4}$: $\cos\theta = \frac{1}{\sqrt{2}}$, $\cos^2\theta = \frac{1}{2}$, $\sin\theta = \frac{1}{\sqrt{2}}$, $\tan^2\theta = 1$.$l_1 = 4\sqrt{2} \cdot \frac{1}{2} = 2\sqrt{2}$, $l_2 = 2\sqrt{6} \cdot \frac{1}{2} = \sqrt{6}$, so $l_1 l_2 = 2\sqrt{2} \cdot \sqrt{6} = 2\sqrt{12} = 4\sqrt{3}$.$e_1 = \sqrt{1 + \cos^2\theta} = \sqrt{1 + \frac{1}{2}} = \frac{\sqrt{3}}{\sqrt{2}}$, $e_2 = \sin\theta = \frac{1}{\sqrt{2}}$, so $e_1 e_2 = \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2}$.Thus $\frac{l_1 l_2}{e_1 e_2} = \frac{4\sqrt{3}}{\sqrt{3}/2} = 8$, and $\frac{l_1 l_2}{e_1 e_2} \cdot \tan^2\theta = 8 \cdot 1 = 8$.Answer:$8$ (Option A)