Question

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

Given equation of hyperbola is
$x^2-3y^2=3$
$\Rightarrow \frac{x^2}{3}-\frac{y^2}{1}=1$
Here, $a^2=3$ and $b^2=1,a>b$
Now, the equation of asymptote of this hyperbola is,
$y=\pm \frac{b}{a}x$
$\Rightarrow y=\pm \frac{1}{\sqrt{3}}x$
$\Rightarrow y=\frac{x}{\sqrt{3}}...(i)$
and $y=\frac{-x}{\sqrt{3}}...(ii)$
Let slope of asymptote (i) is, $m_1=\frac{1}{\sqrt{3}}$
and slope of asymptote (ii) is, $m_2=\frac{-1}{\sqrt{3}}$
Let $\theta$ be the angle between both asymptote, then
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{1/\sqrt{3}+1/\sqrt{3}}{1-1/3}\right|$
$\Rightarrow \tan \theta =\left|\frac{2/\sqrt{3}}{2/3}\right|=\sqrt{3}$
$\Rightarrow \tan \theta =\tan 60^{\circ }$
$\Rightarrow \theta =\pi /3$