• # question_answer If $PQ$ is a double 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 A) $1<e<\frac{2}{\sqrt{3}}$ B) $e=\frac{2}{\sqrt{3}}$ C) $e=\frac{\sqrt{3}}{2}$ D) $e>\frac{2}{\sqrt{3}}$

Let the hyperbola be $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$ and any double ordinate $PO$ be$(a\sec \theta ,\,\,b\tan \theta ),\,\,(a\sec \theta ,\,\,-b\tan \theta )$ and $O$ is centre$(0,\,\,0)$. $\Delta OPQ$being equilateral. $\Rightarrow$               $\tan {{30}^{o}}=\frac{b\tan \theta }{a\sec \theta }$ $\Rightarrow$               $3\frac{{{b}^{2}}}{{{a}^{2}}}=\cos \text{e}{{\text{c}}^{2}}\theta$ $\Rightarrow$               $3({{e}^{2}}-1)=\cos \text{e}{{\text{c}}^{2}}\theta$ Now,     $\cos \text{e}{{\text{c}}^{2}}\theta \ge 1$ $\Rightarrow$               $3({{e}^{2}}-1)\ge 1$ $\Rightarrow$               ${{e}^{2}}\ge 4/3$ $\Rightarrow$               $e>\frac{2}{\sqrt{3}}$