Question

If $PQ$ is a double ordinate of 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. $e=\frac{\sqrt{3}}{2}$
• D. $e>\frac{2}{\sqrt{3}}$

Answer 1

Answer: (D)

Let $P(a \sec \theta, b \tan \theta) ; Q(a \sec \theta,-b \tan \theta)$ be end
points of double ordinates and $(0,0)$ is the centre of the hyperbola.
Now, $P Q=2 b \tan \theta$

$O Q=O P=\sqrt{a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta} $
Since, $ O Q=O P=P Q$
$\therefore 4 b^{2} \tan ^{2} \theta=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta $
$\Rightarrow 3 b^{2} \tan ^{2} \theta=a^{2} \sec ^{2} \theta $
$\Rightarrow 3 b^{2} \sin ^{2} \theta=a^{2} $
$\Rightarrow 3 a^{2}\left(e^{2}-1\right) \sin ^{2} \theta=a^{2} $
$\Rightarrow 3\left(e^{2}-1\right) \sin ^{2} \theta=1 $
$\Rightarrow \frac{1}{3\left(e^{2}-1\right)}=\sin ^{2} \theta < 1, $
$\Rightarrow \frac{1}{e^{2}-1} < 3 $
$\Rightarrow e^{2}-1 > \frac{1}{3} $
$\Rightarrow e^{2} > \frac{4}{3}$
$\therefore e > \frac{2}{\sqrt{3}}$