Question

If the angle between the asymptotes of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{\pi }{3}.$ Then the eccentricity of conjugate hyperbola is

Answer 1

We know that, angle between asymptotes of hyperbola is $2{\tan }^{-1}\left(\frac{b}{a}\right)$
$\therefore 2{\tan }^{-1}\left(\frac{b}{a}\right)=\frac{\pi }{3}$
$\Rightarrow \frac{b}{a}=\tan \left(\frac{\pi }{6}\right)$
$\Rightarrow \frac{b}{a}=\frac{1}{\sqrt{3}}$
Now,
eccentricity, $e=\sqrt{1+\frac{b^2}{a^2}}$
$\therefore e=\sqrt{1+\frac{1}{3}}=\sqrt{\frac{4}{3}}$
Let $e^{\prime }$ be the eccentricity of conjugate ​hyperbola
$\Rightarrow \frac{1}{e^{\prime 2}}+\frac{1}{e^2}=1$
$\Rightarrow \frac{1}{e^{\prime 2}}+\frac{3}{4}=1$
$\Rightarrow \frac{1}{e^{\prime 2}}=\frac{1}{4}$
​​​​​$\therefore e^{\prime }=2$​