• # question_answer If PQ is a double ordinate of 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         [EAMCET 1999] A)            $1<e<2/\sqrt{3}$                     B)            $e=2/\sqrt{3}$ C)            $e=\sqrt{3}/2$                         D)            $e>2/\sqrt{3}$

Correct Answer: D

Solution :

Let P $(a\sec \theta ,\,b\tan \theta );\,Q(a\sec \theta ,\,-b\tan \theta )$ be end points of double ordinates and $C(0,\,0)$, is the centre of the hyperbola. Now $PQ=2b\,\tan \theta$            $CQ=CP=\sqrt{{{a}^{2}}{{\sec }^{2}}\theta +{{b}^{2}}{{\tan }^{2}}\theta }$            Since $CQ=CP=PQ$,            $\therefore \,4{{b}^{2}}{{\tan }^{2}}\theta ={{a}^{2}}{{\sec }^{2}}\theta +{{b}^{2}}{{\tan }^{2}}\theta$            Þ $3{{b}^{2}}{{\tan }^{2}}\theta ={{a}^{2}}{{\sec }^{2}}\theta$ Þ $3{{b}^{2}}{{\sin }^{2}}\theta ={{a}^{2}}$            Þ $3{{a}^{2}}({{e}^{2}}-1){{\sin }^{2}}\theta ={{a}^{2}}$Þ $3({{e}^{2}}-1){{\sin }^{2}}\theta =1$            Þ $\frac{1}{3({{e}^{2}}-1)}={{\sin }^{2}}\theta <1$,  $(\because \,{{\sin }^{2}}\theta <1)$                    $\Rightarrow$ $\frac{1}{{{e}^{2}}-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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