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 :

• A. (2, 3]
• B. $(3, \infty )$
• C. (3/2, 2]
• D. (1, 3/2]

Answer 1

Answer: (B)

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