Question

If e is the eccentricity of the hyperbola and $\theta$ is angle between the asymptotes, then $\frac{cos\theta }{2}$ =

• A. $\frac{(1-e)}{e}$
• B. $\frac{1}{e}-1$
• C. $\frac{1}{e}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{e}$

Let $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be the hyperbola

It has asymptotes $y=\pm \frac{b}{a}x$

Angle between the asymptotes $=2ta{n}^{-}1\left(\frac{b}{a}\right)=\theta$

$\Rightarrow \tan \frac{\theta }{2}=\pm \frac{b}{a}$

$\Rightarrow se{c}^2\frac{\theta }{2}=1+{\tan }^2\frac{\theta }{2}=1+\frac{b^2}{a^2}$

$\Rightarrow {\sec }^2\frac{\theta }{2}=\sqrt{1+\frac{b^2}{a^2}}$

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

$\Rightarrow \cos \frac{\theta }{2}=\frac{1}{e}$