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
$\ell (L.R)\in (3,\infty )$.