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Q. If $e$ is the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $\theta$ is the angle between the asymptotes, then $\cos \frac{\theta}{2}$ is equal to

Conic Sections

Solution:

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