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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $PQ$ be a chord of the hyperbola
$\frac{x^2}{4} - \frac{y^2}{b^2} = 1$,
perpendicular to the x-axis such that triangle $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of triangle $OPQ$ is:
Solution
$e = \sqrt{1 + \frac{b^2}{4}} = \sqrt{3}$
$\Rightarrow b = 8$
Hyperbola
$\frac{x^2}{4} - \frac{y^2}{64} = 1$
$PM/OM = \tan30^\circ$
$\Rightarrow \frac{2\sqrt{2}\tan\theta}{2\sec\theta} = \frac{1}{\sqrt{3}}$
$\Rightarrow \sin\theta = \frac{1}{\sqrt{6}}$
Area
$= 2 \times \frac{1}{2} \times OM \times MP$
$= 2\sec\theta \cdot 2\sqrt{2}\tan\theta$
$= 4\sqrt{2} \cdot \frac{\sin\theta}{\cos^2\theta}$
$= 4\sqrt{2} \cdot \frac{1}{\sqrt{6}\left(1 - \frac{1}{6}\right)}$
$= \frac{8\sqrt{3}}{5}$
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