Question

The angle between the asymptotes of a hyperbola is ${30}^{\circ }$. The eccentricity of the hyperbola may be :

• A. $\sqrt{6}-\sqrt{2}$
• B. $\sqrt{6}\pm \sqrt{2}$
• C. $\sqrt{6}+\sqrt{2}$
• D. $\sqrt{4+2\sqrt{2}}$

Answer 1

Answer: (B)

$\sqrt{6}\pm \sqrt{2}$

Using angle between asymptotes $=2ta{n}^{-1}\frac{b}{a}={30}^{\circ }or\pi -2ta{n}^{-1}\frac{b}{a}={30}^{\circ }$
$\Rightarrow \frac{b}{a}=tan{15}^{\circ }or\frac{b}{a}=tan{75}^{\circ }$
$\Rightarrow e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+ta{n}^2{75}^{\circ }}$ or $\sqrt{1+ta{n}^2{15}^{\circ }}$
$=sec{15}^{\circ }orcosec{15}^{\circ }=\sqrt{6}\pm \sqrt{2}$
Hence, option 'B' is correct.