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. $(1,\frac{3}{2}]$
• B. $(\frac{3}{2},2]$
• C. $(2,3]$
• D. $(3,\infty )$

Answer 1

The correct option is D $(3,\infty )$

Here, $a^2={\cos }^2\theta$ and $b^2={\sin }^2\theta$
Eccentricity, $e=\sqrt{1+{\left(\frac{b}{a}\right)}^2}=\sqrt{1+{\tan }^2\theta }=\sec \theta$
Given, $e>2\therefore \sec \theta >2$
$\Rightarrow \cos \theta <\frac{1}{2}\Rightarrow \theta \in (\frac{\pi }{3},\frac{\pi }{2})$

Let $L$ be the length of the latus rectum.
Then $L=\frac{2b^2}{a}=\frac{2\times {\sin }^2\theta }{\cos \theta }=2\tan \theta .\sin \theta$, which is strictly increasing in $\theta \in (\frac{\pi }{3},\frac{\pi }{2})$.
So, minimum and maximum value of latus rectum occurs at $\theta =\frac{\pi }{3}$ and $\theta =\frac{\pi }{2}$ respectively.
$L_{min}=\frac{2\times {\sin }^2\frac{\pi }{3}}{\cos \frac{\pi }{3}}=3$
$L_{max}=\frac{2\times {\sin }^2\frac{\pi }{2}}{\cos \frac{\pi }{2}}\to \infty$
$\therefore L\in (3,\infty )$