If $PQ$ is a double ordinate of the 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

Question

Dr. Abubacker · Accepted Answer

(b)
$\because PQ$ Is the double ordinate. Let $MP = MQ = l.$
Given that $\mathrm{\Delta}OPQ$ is an equilateral, then $OP = OQ = PQ$
$\Rightarrow (OP)^{2} = (OQ)^{2} = (PQ)^{2}$
$\Rightarrow \frac{a^{2}}{b^{2}}\left( b^{2} + l^{2} ight) + l^{2} = \frac{a^{2}}{b^{2}}\left( b^{2} + l^{2} ight) + l^{2} = 4l^{2}$
$\Rightarrow \frac{a^{2}}{b^{2}}\left( b^{2} + l^{2} ight) + 3l^{2}$
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$\Rightarrow a^{2} = l^{2}\left( 3 - \frac{a^{2}}{b^{2}} ight)$
$\Rightarrow \ \ l^{2} = \frac{a^{2}b^{2}}{\left( 3b^{2} - a^{2} ight)} > 0$
$	herefore\ \ 3b^{2} - a^{2} > 0$
$\Rightarrow \ \ 3b^{2} > a^{2}$
$\Rightarrow 3a^{2}\left( e^{2} - 1 ight) > a^{2}$
$\Rightarrow e^{2} > 4/3$
$	herefore\ \ e > \frac{2}{\sqrt{3}}$

Q. If $PQ$ is a double ordinate of the 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

Explanation

Reviewed by Dr. Abubacker

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