Question

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola, then
(where $e$ is the eccentricity of the hyperbola)

• A. $1<e<\frac{2}{\sqrt{3}}$
• B. $e=\frac{2}{\sqrt{3}}$
• C. $1<e<\frac{3}{2}$
• D. $e>\frac{2}{\sqrt{3}}$

Answer 1

The correct option is D $e>\frac{2}{\sqrt{3}}$

Given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The coordinates of $P$ and $Q$ are
$P\equiv (a\sec \theta ,b\tan \theta )Q\equiv (a\sec \theta ,-b\tan \theta )$
As $△OPQ$ is equilateral, so
$\tan {30}^{\circ }=\frac{b\tan \theta }{a\sec \theta }\Rightarrow \frac{\sqrt{3}b}{a}=\text{cosec}\theta \Rightarrow \frac{3b^2}{a^2}={\text{cosec}}^2\theta \Rightarrow 3(e^2-1)={\text{cosec}}^2\theta \Rightarrow e^2=\frac{{\text{cosec}}^2\theta }{3}+1$
As ${\text{cosec}}^2\theta >1$, so
$e^2>\frac{4}{3}\Rightarrow e>\frac{2}{\sqrt{3}}$