Question

Angle between the asymptotes of a hyperbola is ${30}^{\circ }$ then $e=$

• A. $\sqrt{6}$
• B. $\sqrt{2}$
• C. $\sqrt{6}-\sqrt{2}$
• D. $\sqrt{6}-\sqrt{3}$

Answer 1

The correct option is C $\sqrt{6}-\sqrt{2}$

Asymptotes of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

are $y=\pm \frac{b}{a}x$

Then if angle between them is ${30}^o$

$\tan {30}^o=\frac{\frac{b}{a}+\frac{b}{a}}{1-\frac{b^2}{a^2}}=\frac{2ab}{a^2-b^2}$

$\frac{1}{\sqrt{3}}=\frac{2ab}{a^2-b^2}$

$\frac{1}{\sqrt{3}}=\frac{2\left(b/a\right)}{1-(b/a{)}^2}$

Put $x=b/a$

$\frac{1}{\sqrt{3}}=\frac{2x}{1-x^2}$

$1-x^2=2\sqrt{3}x$

$x^2+2\sqrt{3}x-1=0$

$x=\frac{-2\sqrt{3}\pm \sqrt{16}}{2}$

$=-\sqrt{3}\pm 2$

x can't be negative, so

$x=2-\sqrt{3}$

$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+x^2}$

$e=\sqrt{1+4+3-4\sqrt{3}}$

$=\sqrt{8-4\sqrt{3}}$

$e=2\sqrt{2-\sqrt{3}}$

$e=\sqrt{6}-\sqrt{2}$

C is correct.