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 \[\frac{1}{3({{e}^{2}}-1)}={{\sin }^{2}}\theta <1\],  \[(\because \,{{\sin }^{2}}\theta <1)\]                    \[\Rightarrow \] \[\frac{1}{{{e}^{2}}-1}<3\]\[\Rightarrow \,\,{{e}^{2}}-1>\frac{1}{3}\]\[\Rightarrow \,\,{{e}^{2}}>\frac{4}{3}\]\[\Rightarrow \,e>\frac{2}{\sqrt{3}}\].