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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:

 

  1. (1) $(3, \infty)$

  2. (2) $(3 / 2,2]$

  3. (3) $(2,3]$

  4. (4) $(1,3 / 2]$


Correct Option: 1

Solution:

$\because a^{2}=\cos ^{2} \theta, b^{2}=\sin ^{2} \theta$

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 $(3, \infty)$

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