Ellipse

Curve $x^{2}+2 y^{2}=2$ intersect the line $x+y=1$ at points $P$ and $Q$. First we have to find any common relation between these two curves. Use substitution for the same as follows, $x^{2}+2 y^{2} \;\; =2$ . . . (i) $x+y \;\; =1, \;\text{ then }\;(x+y)^{2}=1^{2}$ $\Rightarrow \;\; x^{2}+y^{2}+2 x y=1$. . . (ii) We can write Eq. (i) as, $x^{2}+2 y^{2}-2(1)^{2}=0$ $\Rightarrow \;\; x^{2}+2 y^{2}-2(x+y)^{2}=0$ $\Rightarrow x^{2}+2 y^{2}-2 x^{2}-2 y^{2}-4 x y=0$ $\Rightarrow \;\; -x^{2}-4 x y=0 \Rightarrow -x(x+4 y)=0$ Gives, $x=0$ and $x+4 y=0$ or $y=\;\frac{-1}{4}x$ Draw the line $y=\;\frac{-1}{4}x$ on graph and take arbitrary point (any one) as follows, From given graph,$\tan \theta = \;\frac{1}{4} \Rightarrow \theta = \tan^{-1}\left(\;\frac{1}{4}\right)$ We have two lines, $y=-\;\frac{1}{4}x$ and $x=0$ (i.e. $Y$ - axis). Thus, any line joining these two curves makes an angle $\;\frac{\pi}{2}+\theta$ at origin. $\therefore$ Answer is $\;\frac{\pi}{2}+\tan^{-1}\left(\;\frac{1}{4}\right)$.