Let $0<\theta<\frac{\pi}{2}$. If the eccentricity of the hyperbola
$\frac{x^{2}}{\cos ^{2} \theta}-\frac{y^{2}}{\sin ^{2} \theta}=1$ is greater than 2 , then the length of
its latus rectum lies in the interval:
Correct Option: 1
$\because a^{2}=\cos ^{2} \theta, b^{2}=\sin ^{2} \theta$
and $e>2 \Rightarrow e^{2}>4 \Rightarrow 1+b^{2} / a^{2}>4$
$\Rightarrow \quad 1+\tan ^{2} \theta>4$
$\Rightarrow \sec ^{2} \theta>4 \Rightarrow \theta \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
Latus rectum,
$L R=\frac{2 b^{2}}{a}=\frac{2 \sin ^{2} \theta}{\cos \theta}=2(\sec \theta-\cos \theta)$
$\Rightarrow \quad \frac{d(L R)}{d \theta}=2(\sec \theta \tan \theta+\sin \theta)>0 \forall \theta \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$
$\therefore \quad \min (L R)=2\left(\sec \frac{\pi}{3}-\cos \frac{\pi}{3}\right)=2\left(2-\frac{1}{2}\right)=3$
$\max (L R)$ tends to infinity as $\theta \rightarrow \frac{\pi}{2}$
Hence, length of latus rectum lies in the interval $(3, \infty)$