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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 center of the hyperbola. then the $'e'$ eccentricity of the hyperbola, satisfies
$1\, < \,e\,<\,\frac{2}{{\sqrt 3 }}$
$e\, = \,\frac{2}{{\sqrt 3 }}$
$e\, = \,\frac{{\sqrt 3 }}{2}$
$e\, > \,\frac{2}{{\sqrt 3 }}$
Solution
Let the coordinates of $P$ be $(\alpha, \beta)$ Then, $\mathrm{PQ}=2 \beta$ and $\mathrm{OP}=\sqrt{\alpha^{2}+\beta^{2}}$
Since $OPQ$ is an equilateral triangle $OP = PQ$
$\Rightarrow \quad \alpha^{2}+\beta^{2}=4 \beta^{2} $
$\Rightarrow \alpha^{2}=3 \beta^{2} \quad \Rightarrow \quad \alpha=\pm \sqrt{3} \beta$
Also since $(\alpha, \beta)$ lies on the given hyperbola, $\frac{\alpha^{2}}{a^{2}}-\frac{\beta^{2}}{b^{2}}=1$
$\Rightarrow \quad \frac{3 \beta^{2}}{a^{2}}-\frac{\beta^{2}}{b^{2}}=1 $
$\Rightarrow \frac{3}{a^{2}}-\frac{1}{b^{2}}=\frac{1}{\beta^{2}}>0$
$ \Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} $
$\Rightarrow e^{2}-1>\frac{1}{3}$
$\Rightarrow \quad e^{2}>\frac{4}{3} $
$\Rightarrow e>\frac{2}{\sqrt{3}}$
