Let 0 < θ < π/2 . If the eccentricity of the


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:

  1. $(2,3]$

  2. $(3, \infty)$

  3. $(3 / 2,2]$

  4. $(1,3 / 2]$

Correct Option: , 2


$\mathrm{e}=\sqrt{1+\tan ^{2} \theta}=\sec \theta$

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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