Question

If $AB$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $△ABC$ is equilateral, $C$ being centre of hyperbola, then eccentricity $e$ of hyperbola satisfies

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

Answer 1

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


$AB=2b\tan \theta$
and $AC=\sqrt{a^2{\sec }^2\theta +b^2{\tan }^2\theta }$
Since $△ABC$ is equilateral,
therefore, $AC=AB$
$\Rightarrow a^2{\sec }^2\theta +b^2{\tan }^2\theta =4b^2{\tan }^2\theta$
$\Rightarrow a^2{\sec }^2\theta =3b^2{\tan }^2\theta$
$\Rightarrow \frac{1}{3(e^2-1)}={\sin }^2\theta (\because b^2=a^2(e^2-1\left)\right)$
$\Rightarrow e^2-1>\frac{1}{3}(\because {\sin }^2\theta <1)$
$\therefore e>\frac{2}{\sqrt{3}}$