Question

If $P Q$ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that $O P Q$ 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 $P Q$ be $(a \sec \theta, b \tan \theta),(a \sec \theta,-b \tan \theta)$ and $O$ is centre $(0,0)$

$\Delta O P Q$ 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, $\text{cosec}^{2} \theta \geq 1$
$\therefore 3\left(e^{2}-1\right) \geq 1 $
$\Rightarrow e^{2} \geq \frac{4}{3}$
$\Rightarrow e >2 / \sqrt{3}$