Mathematics | VITEEE 2014 | Discussion Page

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

$\tan^{-1}\frac{2}{\sqrt{3}}$

$\frac{\pi}{6}$

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

$\frac{\pi}{2}$

Option C

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

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

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