Question:

<p><strong>Step 1: Identify the hyperbola parameters.</strong></p><p>The hyperbola is $\frac{x^2}{\cos^2\theta} - \frac{y^2}{\sin^2\theta} = 1$</p><p>Here: $a^2 = \cos^2\theta$ and $b^2 = \sin^2\theta$</p><p>So $a = |\cos\theta|$ and $b = |\sin\theta|$ (taking positive values)</p><p><strong>Step 2: Express eccentricity in terms of θ.</strong></p><p>For a hyperbola: $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{\sin^2\theta}{\cos^2\theta} = 1 + \tan^2\theta = \sec^2\theta$</p><p>Therefore: $e = |\sec\theta| = \frac{1}{|\cos\theta|}$</p><p><strong>Step 3: Apply the constraint e > 2.</strong></p><p>$\frac{1}{|\cos\theta|} > 2$</p><p>$|\cos\theta| < \frac{1}{2}$</p><p>Since $a^2 = \cos^2\theta$, we have: $\cos^2\theta < \frac{1}{4}$</p><p>This gives: $\sin^2\theta > \frac{3}{4}$ (using $\sin^2\theta + \cos^2\theta = 1$)</p><p><strong>Step 4: Find the latus rectum.</strong></p><p>Latus rectum: $L = \frac{2b^2}{a} = \frac{2\sin^2\theta}{|\cos\theta|}$</p><p>Since $e = \frac{1}{|\cos\theta|}$, we have $|\cos\theta| = \frac{1}{e}$</p><p>Therefore: $L = 2\sin^2\theta \cdot e = 2(1 - \cos^2\theta) \cdot e$</p><p><strong>Step 5: Express L in terms of e.</strong></p><p>From $e = \frac{1}{|\cos\theta|}$, we get $\cos^2\theta = \frac{1}{e^2}$</p><p>Thus: $L = 2\left(1 - \frac{1}{e^2}\right) \cdot e = 2e - \frac{2}{e}$</p><p><strong>Step 6: Find the range of L when e > 2.</strong></p><p>Let $f(e) = 2e - \frac{2}{e}$ for $e > 2$</p><p>$f'(e) = 2 + \frac{2}{e^2} > 0$ for all $e > 0$, so $f$ is strictly increasing.</p><p>At $e = 2$: $f(2) = 4 - 1 = 3$</p><p>As $e \to \infty$: $f(e) \to \infty$</p><p>Since $f$ is strictly increasing and continuous on $(2, \infty)$, the range is $(3, \infty)$</p><p><strong>∴ Answer: A</strong></p>