Question:
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: , 2
Solution:
$\mathrm{e}=\sqrt{1+\tan ^{2} \theta}=\sec \theta$
As, $\sec \theta>2 \Rightarrow \cos \theta<\frac{1}{2}$
$\Rightarrow \theta \in\left(60^{\circ}, 90^{\circ}\right)$
Now, $\ell(\mathrm{L} \cdot \mathrm{R})=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2 \frac{\left(1-\cos ^{2} \theta\right)}{\cos \theta}$
$=2(\sec \theta-\cos \theta)$
Which is strictly increasing, so
$\ell$ (L.R) $\in(3, \infty)$
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