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If e is the eccentricity of the hyperbol...

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

A

`(1-e)/(e)`

B

`(2)/(e ) -e`

C

`(1)/(e )`

D

`(2)/(e )`

Text Solution

Verified by Experts

The correct Answer is:
C
`theta = 2 tan^(-1).(b)/(a) rArr tan.(theta)/(2) = (b)/(a)`
`rArr cos.(theta)/(2) = (a)/(sqrt(a^(2)+b^(2))) = (1)/(sqrt(1+(b^(2))/(a^(2)))) = (1)/(e)`
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