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. $\left(\frac{3}{2},2\right]$
• D. $\left(1,\frac{3}{2}\right]$

Answer 1

Answer: (B)

$(3,\infty )$

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

As, $\sec \theta >2\Rightarrow \cos \theta <\frac{1}{2}$

$\Rightarrow \theta ϵ({60}^{\circ },{90}^{\circ })$

Now, $l(L.R)=\frac{2b^2}{a}=2\frac{(1-{\cos }^2\theta )}{\cos \theta }$

$=2(\sec \theta -\cos \theta )$

Which is strictly increasing, so

$l(L.R)ϵ(3,\infty )$.