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Question

If PQ is a double ordinate of hyperbola x2a2y2b2=1 such that OPQ is an equilateral triangle, O being the center of the hyperbola. Then, the eccentricity e is the hyperbola satisfies :

A
1<e<23
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B
e=23
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C
e=32
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D
e>23
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Solution

The correct option is D e>23
Let P(asecθ,btanθ);Q(asecθ,btanθ) be end points of double ordinate and C(0,0) is the centre of the hyperbola.
Now, PQ=2btanθ
CQ=CP=a2sec2θ+b2tan2θ
since CQ=CP=PQ
4b2tan2θ=a2sec2θ+b2tan2θ3b2tan2θ=a2sec2θ
3b2sin2θ=a23a2(e21)sin2θ=a2
3(e21)sin2θ=113(e21)=sin2θ<1 (sin2θ<1)
1e21<3e21>13

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