Q.

If the eccentricity of a hyperbola is 3/2 then the angle between asymptotes of conjugate hyperbola is 
 

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a

cos-1(1/3)

b

cos-1(2/3)

c

π/2

d

cos-1(1/9)

answer is B.

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Detailed Solution

Given e=32

we know that  1e2+1e12=1 1e12=1-49 e12=95 e1=35

let 'θ' be angle between asymptotes of conjugate hyperbola

θ=2sec-1(e1) sec(θ2)=e1 cosθ2=1e1 cosθ2=53 Now cosθ=2cos2θ2-1                   =259-1          cosθ=19 θ=cos-119

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