Question
Let PQ be a chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{{\text{ }b}^{2}} = 1$, perpendicular to the x -axis such that OPQ is an equilateral triangle, $O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of the triangle OPQ is :
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Solution
$e = \sqrt{1 + \frac{b}{4}} = \sqrt{3}\ \Rightarrow \text{ }b = 8$
∴ Hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{8} = 1$
$${\frac{PM}{OM} = tan30^{\circ} }{\Rightarrow \frac{2\sqrt{2}tan\theta}{2sec\theta} = \frac{1}{\sqrt{3}} \Rightarrow sin\theta = \frac{1}{\sqrt{6}} }$$Area $= 2 \times \frac{1}{2} \times OM \times MP$
$${= 2sec\theta \times 2\sqrt{2}tan\theta }{= 4\sqrt{2}\frac{sin\theta}{\cos^{2}\theta} = 4\sqrt{2} \times \frac{1}{\sqrt{6} \times \left( 1 - \frac{1}{6} \right)} = \frac{8\sqrt{3}}{5}}$$
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