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, $O$ bing the centre of the hyperbola, then the eccentricity $e$ 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)

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 $(0,0)$

$\Delta OPQ$ being equilateral.
$\therefore \tan 30^{\circ }=\frac{b\tan \theta }{a\sec \theta }$
$\Rightarrow 3\cdot \frac{b^2}{a^2}={\text{cosce}}^2\theta$
$\Rightarrow 3\left(e^2-1\right)={\text{cosce}}^2\theta$
Now, ${\text{cosec}}^2\theta \ge 1$
$\therefore 3\left(e^2-1\right)\ge 1$
$\Rightarrow e^2\ge \frac{4}{3}$
$\Rightarrow e>2/\sqrt{3}$