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If e is the eccentricity of x^2/a^2-y^2/...

If `e` is the eccentricity of `x^2/a^2-y^2/b^2=1` and `theta` be the angle between its asymptotes, then `cos(theta/2)` is equal to

A

`e=sqrt((a^2+b^2)/(a^2))`

B

`e=sqrt((a^2-b^2)/(a^2))`

C

`e=sqrt((b^2-a^2)/(a^2))`

D

`e=sqrt((a^2+b^2)/(b^2))`

Text Solution

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The correct Answer is:
D
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