Question

Show that angle between the two asymptotes of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2{\tan }^{-1}\left(\frac{b}{a}\right)$ or $2{\sec }^{-1}\left(e\right)$.

Answer 1

Asymptotes of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are given by $\frac{x^2}{a}-\frac{y^2}{b^2}=0$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=0$ and $\frac{x}{a}-\frac{y}{b}=0$

As slope $=\frac{-A}{B}$

$\Rightarrow m_1=\frac{-\frac{1}{a}}{\frac{1}{b}}=\frac{-b}{a}$, $m_2=\frac{b}{a}$, let $\theta$ be angle between lines

$\Rightarrow \tan \theta =\frac{m_2-m_1}{1+m_1{m}_2}=\frac{\frac{2b}{a}}{1-\frac{b^2}{a^2}}$

$\Rightarrow \theta 2{\tan }^{-1}\left(\frac{b}{a}\right)$

$\{$ As $2{\tan }^{-1}x=\tan \left(\frac{2x}{1-x^2}\right)\}$

Now, ${\tan }^{-1}\frac{x}{y}={\sec }^{-1}\frac{\sqrt{x^2+y^2}}{y}\{{\tan }^{-1}\frac{P}{B},{\sec }^{-1}\frac{H}{B}\}$

$\Rightarrow 2{\tan }^{-1}\left(\frac{b}{a}\right)=2{\sec }^{-1}\left(\frac{\sqrt{a^2+b^2}}{a}\right)$ and $a^2+b^2=a^2{e}^2$

$=2{\sec }^{-1}\left(e\right)$.