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Question

Mathematics Question on Hyperbola

If PQPQ is a double ordinate of the hyperbola x2a2y2b2=1\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 such that ΔOPQ\Delta OPQ is equilateral, OO being the centre. Then the eccentricity e satisfies

A

1<e<231 < e < \frac{2}{\sqrt{3}}

B

e=22e=\frac{2}{\sqrt{2}}

C

e=32e=\frac{\sqrt{3}}{2}

D

e>23e>\frac{2}{\sqrt{3}}

Answer

e>23e>\frac{2}{\sqrt{3}}

Explanation

Solution

ΔOPQ\because \Delta OPQ is equilateral, OP=PQOP = PQ a2sec2θ+b2tan2θ=(2btanθ)2\Rightarrow a^{2}\,sec^{2}\,\theta+b^{2}\,tan^{2}\,\theta=\left(2b\,tan\,\theta\right)^{2} a2sec2θ+3b2tan2θ\Rightarrow a^{2}\,sec^{2}\,\theta+3b^{2}\,tan^{2}\,\theta sin2θ=a23b2\Rightarrow sin^{2}\,\theta= \frac{a^{2}}{3b^{2}} Now, sin2θ<1sin^{2}\,\theta <1 a23b2>1\Rightarrow \frac{a^{2}}{3b^{2}} >1 b2a2>131+b2a2>43e2>43e>23\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} \Rightarrow 1+\frac{b^{2}}{a^{2}} > \frac{4}{3} \Rightarrow e^{2}>\frac{4}{3} \Rightarrow e> \frac{2}{\sqrt{3}}