Question

The angle between lines joining the origin to the points of intersection of the line $\sqrt{3}x+y=2$ and the curve $y^2-x^2=4$ is

• A. ${\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$
• B. $\frac{\pi }{6}$
• C. ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Making the curve $x^2-y^2=4$ homogeneous with help of straight line $\sqrt{3x}+y=2$ we get

$y^2-x^2=4{\left(\frac{\sqrt{3}x+y}{2}\right)}^2$

$\Rightarrow 4x^2+2\sqrt{3}xy=0$

If $\theta$ is the angle between these lines ,then

$\theta ={\tan }^{-1}\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Here, $a=4,b=0,h=\sqrt{3}$

$\Rightarrow \theta ={\tan }^{-1}\left|\frac{2\sqrt{3-0}}{4+0}\right|$

$\theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$