Question

If $e$ is the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\theta$ be the angle between the asymptotes then $\sec \frac{\theta }{2}$ equals :

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

Answer 1

The correct option is D $e$

Equation of asymptotes to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are given by

$y=-\frac{b}{a}x$

$\therefore m_1=-\frac{b}{a}$

Similarly $y=\frac{bx}{a}$

$\therefore m_2=\frac{b}{a}$

Now $\theta =2{\tan }^{-1}\frac{b}{a}$

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

$\Rightarrow {\sec }^2\frac{\theta }{2}=e^2\Rightarrow \sec \frac{\theta }{2}=e$