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 $ ax^{2} + 2 h xy +by^{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} $
$\theta= \tan^{-1} \left(\frac{\sqrt{3}}{2}\right)$