$AB$ is double ordinate of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ such that $

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$AB$ is double ordinate of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ such that $\bigtriangleup AOB$ (where ' $O$ ' is the origin) is an equilateral triangle, then the eccentricity $e$ of the hyperbola satisfies :

Accepted Answer

Sol. (d)Put $\left( l\frac{\sqrt{3}}{2},\frac{l}{2} \right)$ to equation of hyperbola$${\Rightarrow l^{2}\left( \frac{3}{4a^{2}} - \frac{1}{4b^{2}} \right) = 1 }{\Rightarrow \frac{1}{l} = \frac{3}{4a^{2}} - \frac{1}{4b^{2}} > 0 }{\Rightarrow \frac{b^{2}}{a^{2}} > \frac{1}{3} \Rightarrow e^{2} - 1 > \frac{1}{3} \Rightarrow e^{2} > \frac{4}{3} }{\Rightarrow e > \frac{2}{\sqrt{3}}}$$

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