Question

If $AB$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $\Delta AOB(O$ is the origin$)$ is an equilateral triangle, then the eccentricity $e$ of the hyperbola satisfies :

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

Answer 1

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

Let $O=(0,0),A=(a\sec \theta ,b\tan \theta ),B=(a\sec \theta ,-b\tan \theta )$ since $AB$ is a double ordinate.
Since $\Delta ABO$ is equilateral, $OA=AB$
$\Rightarrow a^2{\sec }^2\theta +b^2{\tan }^2\theta =4b^2{\tan }^2\theta$
$\Rightarrow a^2+a^2{\tan }^2\theta =3b^2{\tan }^2\theta$
$\Rightarrow \frac{b^2}{a^2}=\frac{1+{\tan }^2\theta }{3{\tan }^2\theta }$
$e^2=1+\frac{b^2}{a^2}=1+\frac{1+{\tan }^2\theta }{3{\tan }^2\theta }=\frac{1+4{\tan }^2\theta }{3{\tan }^2\theta }=1+\frac{1}{3{\sin }^2\theta }$

Using ${\sin }^2\theta <1$, we get $e^2>\frac{4}{3}$

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