Question

If $e$ is the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\theta$ is angle between the asymptotes, then $\cos \left(\frac{\theta }{2}\right)=$

• A. $\frac{1}{e}$
• B. $\frac{1}{2e}$
• C. $e^{-2}$
• D. 0

Answer 1

The correct option is A $\frac{1}{e}$

We know, if $\theta$ is the angle between asymptotes then,

$\tan \frac{\theta }{2}=\frac{b}{a}$

$\therefore \cos \frac{\theta }{2}=\frac{a}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2{e}^2}}=\frac{1}{e}$

Using $b^2=a^2(e^2-1)$