<p>Given the equation of hyperbola \(\dfrac{x^2}{\cos^2\theta} - \dfrac{y^2}{\sin^2\theta} = 1\) whose eccentricity \(e > 2\). Then the latus rectum lies in the interval:</p>
Step-by-Step Solution
Key Concept: For a hyperbola, use the relation e² = 1 + b²/a² and apply the constraint e > 2 to find bounds on b²/a², then calculate latus rectum = 2b²/a using the parametric forms a² = cos²θ and b² = sin²θ.
<p><strong>Step 1:</strong> For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, identify $a^2 = \cos^2\theta$ and $b^2 = \sin^2\theta$.</p><p><strong>Step 2:</strong> The eccentricity is $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \tan^2\theta} = |\sec\theta|$.</p><p><strong>Step 3:</strong> Given $e > 2$, we have $|\sec\theta| > 2$, which means $\frac{1}{|\cos\theta|} > 2$, so $\cos^2\theta < \frac{1}{4}$.</p><p><strong>Step 4:</strong> Since $\cos^2\theta + \sin^2\theta = 1$, we get $\sin^2\theta > \frac{3}{4}$.</p><p><strong>Step 5:</strong> The latus rectum is $L = \frac{2b^2}{a} = \frac{2\sin^2\theta}{\cos\theta}$. Since $\sin^2\theta > \frac{3}{4}$ and $\cos^2\theta < \frac{1}{4}$, we have $L = 2\sin^2\theta/\cos\theta > 2 \cdot \frac{3}{4} / \frac{1}{2} = 3$.</p><p><strong>Step 6:</strong> As $\cos^2\theta \to 0^+$, the latus rectum approaches infinity. Therefore, $L \in (3, \infty)$ or the given interval.</p><p>∴ Answer: B</p>
Correct Answer: B