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Question

If AB is a double ordinate of the hyperbola x2a2y2b2=1 such that ABC is equilateral, C being centre of hyperbola, then eccentricity e of hyperbola satisfies

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

The correct option is D e>23

AB=2btanθ
and AC=a2sec2θ+b2tan2θ
Since ABC is equilateral,
therefore, AC=AB
a2sec2θ+b2tan2θ=4b2tan2θ
a2sec2θ=3b2tan2θ
13(e21)=sin2θ (b2=a2(e21))
e21>13 (sin2θ<1)
e>23

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