Question

If the curve $x^2+2y^2=2$ intersects the line $x+y=1$ at two points $P$ and $Q$, then the angle subtended by the line segment $PQ$ at the origin is :

• A. $\frac{\pi }{2}+{\tan }^{-1}\left(\frac{1}{4}\right)$
• B. $\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{4}\right)$
• C. $\frac{\pi }{2}+{\tan }^{-1}\left(\frac{1}{3}\right)$
• D. $\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is A

$\frac{\pi }{2}+{\tan }^{-1}\left(\frac{1}{4}\right)$

Explanation for the correct option.

Step 1: Find the equations of line

Given, that the curve $x^2+2y^2=2$ intersects the line $x+y=1$ at two points $P$ and $Q$.

$$\begin{gathered} x^2+2y^2-2{\left(1\right)}^2=0 \\ {x}^2+2y^2-2(x+y{)}^2=0 \\ \Rightarrow x^2+4xy=0 \\ \Rightarrow x\left(x+4y\right)=0 \\ \Rightarrow x=0,x=-4y \end{gathered}$$

Step 2: Find the angle.

For$a{x}^2+2hxy+b{y}^2=0$, obtuse angle between lines $\theta$ is $\tan \theta =\pm \frac{2\sqrt{h^2-ab}}{a+b}$
$$\begin{gathered} \Rightarrow \tan \theta =\pm 4 \\ \Rightarrow \tan \theta =-4 \\ \Rightarrow \cot \theta =-\frac{1}{4} \\ \Rightarrow \theta ={\cot }^{-1}\frac{1}{4} \\ \Rightarrow \theta =\pi -{\cot }^{-1}\frac{1}{4} \\ \Rightarrow \theta =\pi -\left(\frac{\pi }{2}-{\tan }^{-1}\frac{1}{4}\right) \\ \mathbf{\Rightarrow }\theta =\frac{\pi }{2}+{\tan }^{-1}\frac{1}{4}\Rightarrow \end{gathered}$$
Hence, option A is correct.