Answer:
Option C
Explanation:
On homogenising $y^{2}-x^{2}=4$ with the help of the line $\sqrt{3}x+y=2$, we get
$y^{2}-x^{2}=4\frac{\left(\sqrt{3}x+y\right)^{2}}{4}$
$\Rightarrow$ $y^{2}-x^{2}=3x^{2}+y^{2}+2\sqrt{3}xy$
$\Rightarrow$ $4x^{2}+2\sqrt{3} xy=0$
On comparing with $ax^{2}+2hxy+by^{2}=0$ .
we get,
$a=4,h=\sqrt{3}$ and $b=0$
We know that,
$\tan \theta= 2\frac{\sqrt{h^{2}-ab}}{a+b}=\frac{2\sqrt{3-0}}{4+0}$
$\therefore$ The angle between the lines is
$\theta$=$\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)$
