Question

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $\Delta OPQ$ is equilateral, $O$ being the centre. Then the eccentricity $e$ satisfies:

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

Answer 1

Answer: (D)

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

Let point $P$ be $(h,k)$

Since $PQ$ is a double ordinate, $l\left(PQ\right)=2k$

Also, $l\left(OP\right)=\sqrt{h^2+k^2}$

Since $\Delta OPQ$ is equilateral, $2k=\sqrt{h^2+k^2}$

$\Rightarrow 4k^2=h^2+k^2$ or $h^2=3k^2$

Also, point $P(h,k)$ lies on the hyperbola and so it satisfies $\frac{h^2}{a^2}-\frac{k^2}{b^2}=1$

$\Rightarrow \frac{3k^2}{a^2}-\frac{k^2}{b^2}=1$

$\therefore \frac{3}{a^2}-\frac{1}{b^2}=\frac{1}{k^2}>0$

$\therefore \frac{b^2}{a^2}>\frac{1}{3}$

$e^2=1+\frac{b^2}{a^2}>1+\frac{1}{3}>\frac{4}{3}$ where $e$ is the eccentricity.

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