Q. The angle between lines joining the origin to the point of intersection of the line $\sqrt{3x + y = 2}$ 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) $\frac{\pi}{2}$
Explanation
(c)
On homogenising $y^{2} - x^{2} = 4$ with the help of the line $\sqrt{3}x + y = 2$, we get
$y^{2} - x^{2} = 4\ \frac{\left( \sqrt{3}x + y \right)^{2}}{4}$
$\Rightarrow \ \ y^{2} - x^{2} = 3x^{2} + y^{2} + 2\sqrt{3}xy$
$\Rightarrow 4x^{2} + 2\sqrt{3}xy = 0$
$\therefore\ \tan{\theta = 2\frac{\sqrt{h^{2} - ab}}{a + b}}$
$\Rightarrow \tan{\theta = \frac{2\sqrt{3 - 0}}{4}}$
$\Rightarrow \ \ \theta = \tan^{- 1}\left( \frac{\sqrt{3}}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.