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 is at the origin is

Equation of the curve is $x^2+2y^2=2\Rightarrow \frac{x^2}{2}+\frac{y^2}{1}=1$

            Equation of the line is  x  + y = 1   ; y = 1-x

            Solving the above two equations

$\begin{array}{c}{\text{\hspace{0.17em}x}}^2\text{+2}{\left(1-x\right)}^2\text{=2}\\ x^2+2x^2+2-4x=2\\ 3x^2-4x=0\\ x\left(3x-4\right)=0\end{array}$

Hence the values of x are 0,$\frac{4}{3}$

Hence, the points are  $P(0,1),Q\left(\frac{4}{3},\frac{-1}{3}\right)$

The line OP is vertical axis, the slope of OQ is $\frac{-1}{4}$hence the angle subtended by OP and OQ at origin is  $\theta = \frac{\pi }{2}+\left|{\tan }^{-1}(-\frac{1}{4})\right|$

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