Q.If the angle between the asymptotes is $30^\circ$ then find its eccentricity of hyperbola.

If $2\theta$ is the angle between the asymptotes of a hyperbola, its eccentricity is $e=\sec\theta$; here $2\theta=30^\circ$ gives $e=\sec 15^\circ=\sqrt6-\sqrt2$.

For $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, the asymptotes are $y=\pm\dfrac{b}{a}x$. If $2\theta$ is the angle between them (the angle containing the transverse axis), then $\tan\theta=\dfrac{b}{a}$.

Since $e^2 = 1+\dfrac{b^2}{a^2} = 1+\tan^2\theta = \sec^2\theta$, we get $e=\sec\theta$.

Given the angle between the asymptotes is $30^\circ$, so $2\theta=30^\circ \Rightarrow \theta=15^\circ$.

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