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)

$\because \Delta OPQ$ is equilateral,
$OP=PQ$
$\Rightarrow a^{2}se{c}^2\theta +b^{2}ta{n}^2\theta ={\left(2btan\theta \right)}^2$
$\Rightarrow a^{2}se{c}^2\theta +3b^{2}ta{n}^2\theta$
$\Rightarrow si{n}^2\theta =\frac{a^2}{3b^2}$
Now, $si{n}^2\theta <1$
$\Rightarrow \frac{a^2}{3b^2}>1$
$\Rightarrow \frac{b^2}{a^2}>\frac{1}{3}\Rightarrow 1+\frac{b^2}{a^2}>\frac{4}{3}\Rightarrow e^2>\frac{4}{3}\Rightarrow e>\frac{2}{\sqrt{3}}$