Question

If $PQ$ is a double 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 the eccentricity ${}^{\prime }{e}^{\prime }$ of the hyperbola satisfies.

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

Answer 1

Answer: (D)

$e>\frac{2}{\sqrt{3}}$

Let the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and any double ordinate $PQ$ be $(a\sec \theta ,b\tan \theta ),(a\sec \theta -b\tan \theta )$ and $O$ is centre $(0,0)$.
$△OPQ$ being equilateral
$\Rightarrow \tan {30}^{\circ }=\frac{b\tan \theta }{a\sec \theta }$
$\Rightarrow 3\frac{b^2}{a^2}={\text{cosec}}^2\theta$
$\Rightarrow 3(e^2-1)={\text{cosec}}^2\theta$
Now, ${\text{cosec}}^2\theta \ge 1$
$\Rightarrow 3(e^2-1)\ge 1$
$\Rightarrow e^2\ge 4/3$
$\Rightarrow e>\frac{2}{\sqrt{3}}$