Question

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

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

Answer 1

Answer: (C)

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

On homogenising $y^2-x^2=4$ with the help of the line $\sqrt{3}x+y=2,$ we get

$y^2-x^2=4\frac{2^2}{4}$

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

$\Rightarrow y^2-x^2=3x^2+y^2+2\sqrt{3}xy$

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

On comparing with $a{x}^2+2hxy+by=0$, 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

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

Hence, option C is correct.