Question

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

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

Answer 1

Answer: (C)

On homogenising $y^2-x^2=4$ with the help of the line
$\sqrt{2}x+y=2,$ 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 $a{x}^2+2hxy+b{y}^2=0$ we get
$a=4,h=\sqrt{3}$ and $b=0$
$\therefore$ The angle between the lines is
$\tan \theta =2\frac{\sqrt{h^2-ab}}{a+b}=\frac{2.\sqrt{3-0}}{4+0}$
$\theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$