Answer: (1)

Solution

$a^2 = \cos^2\theta$, $b^2 = \sin^2\theta$. $e^2 = 1 + \tan^2\theta = \sec^2\theta$. $e > 2 \Rightarrow \sec^2\theta > 4 \Rightarrow \theta \in \left(\frac{\pi}{3}, \frac{\pi}{2}\right)$. Latus rectum $= \frac{2b^2}{a} = \frac{2\sin^2\theta}{\cos\theta} = 2(\sec\theta - \cos\theta)$. Its derivative $\frac{d(LR)}{d\theta} = 2(\sec\theta\tan\theta + \sin\theta) > 0$ for $\theta \in (\frac{\pi}{3}, \frac{\pi}{2})$. Min at $\theta = \frac{\pi}{3}$: $LR = 2(\sec\frac{\pi}{3} - \cos\frac{\pi}{3}) = 2(2 - \frac{1}{2}) = 3$. As $\theta \to \frac{\pi}{2}$, $LR \to \infty$. So latus rectum $\in (3, \infty)$.