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 the eccentricity $e$ of the hyperbola satisfies

• A. $1<e<\frac{2}{\sqrt{3}}$
• B. $e>\frac{2}{\sqrt{3}}$
• C. $e=\frac{2}{\sqrt{3}}$
• D. $e>\frac{4}{\sqrt{3}}$

Answer 1

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

Let the coordinates of $P$ be $(h,k)$ and coordinates of $Q$ be $(h,-k)$.
$O(0,0)$ is the centre.
$△OPQ$ is equilateral.
$\therefore O{P}^2=O{Q}^2=P{Q}^2\Rightarrow h^2+k^2=4k^2\Rightarrow h^2=3k^2$

Also, $(h,k)$ lies on the hyperbola.
$\therefore \frac{h^2}{a^2}-\frac{k^2}{b^2}=1\Rightarrow \frac{3k^2}{a^2}-\frac{k^2}{b^2}=1\Rightarrow k^2=\frac{a^2{b}^2}{3b^2-a^2}>0\Rightarrow 3b^2-a^2>0\Rightarrow \frac{b^2}{a^2}>\frac{1}{3}$

Now, $e^2=1+\frac{b^2}{a^2}$
$\Rightarrow e^2>\frac{4}{3}$
$\therefore e>\frac{2}{\sqrt{3}}$