Question

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

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

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

$\text{solve: Given}$

$\sqrt{3}x+y=2$

$\text{and}{y}^2-x^2=4$

$\text{To find the angle between the first}$

$\text{we have to find the equation of another}$

$\text{line.}$

$(0,0)\text{is not satisfy the curve}$

So, we have to homogenise it to get

pair of straight line.

Given eqn. of line is

$\sqrt{3}x+y=2$

$=>{\left(\frac{\sqrt{3}x+y}{2}\right)}^2=1^2$

from eqn. of curve we get

$y^2-x^2=4{\left[\frac{\sqrt{3}x+y}{2}\right]}^2\Rightarrow y^2-x^2=3x^2+y^2+2\sqrt{3}xy$

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

$\Rightarrow 2x[2x+\sqrt{3}y]=0$

$l_1:2x=0\Rightarrow x=0$

slope $=\tan \frac{\pi }{2}$

angle with $x$ axis $=\frac{\pi }{2}$

$l_2:2x+\sqrt{3}y=0y=\frac{-2}{\sqrt{3}}x\text{Slope}=\frac{-2}{\sqrt{3}}$

$\text{slepe of line}{l}_2=\frac{-2}{\sqrt{3}}$

$\Rightarrow \tan \theta =\frac{-2}{\sqrt{3}}$

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

$\text{So, the angle between line}=\frac{\pi }{2}-ta{n}^{-1}\frac{2}{\sqrt{3}}$

$\text{we know that}{\cot }^{-1}x+{\tan }^{-1}x=\frac{\pi }{2}$

$\Rightarrow \frac{\pi }{2}-{\tan }^{-1}x={\cot }^{-1}x$

$\Rightarrow \frac{\pi }{2}-{\tan }^{-1}\frac{2}{\sqrt{3}}={\cot }^{-1}\frac{2}{\sqrt{3}}$

$\text{and}{\tan }^{-1}x={\cot }^{-1}\left(\frac{1}{x}\right)$

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

So, ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is the required angle