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

Given,

$$x^2+2y^2=2 \quad\cdots(1)$$ and $$x+y=1 \quad\cdots(2)$$ 

Let the points of intersection be $$P$$ and $$Q.$$

From (2),

$$y=1-x$$

Substitute into (1):

$$x^2+2(1-x)^2=2$$

$$x^2+2(1-2x+x^2)=2$$

$$3x^2-4x=0$$

$$x(3x-4)=0$$

Hence,

$$x=0\quad\text{or}\quad x=\frac43$$

Therefore,

$$P=(0,1)$$ and $$Q=\left(\frac43,-\frac13\right)$$

Now slopes of

$$OP$$ and $$OQ$$ are $$m_1=\infty$$ and $$m_2=\frac{-1/3}{4/3}=-\frac14$$

Hence,

$$OQ$$ makes an angle $$-\tan^{-1}\left(\frac14\right)$$ with the positive $$x$$-axis.

Since

$$OP$$ lies along the positive $$y$$-axis, it makes angle $$\frac\pi2$$ with the positive $$x$$-axis.

Therefore angle between $$OP$$ and $$OQ$$ is $$\frac\pi2-\left(-\tan^{-1}\left(\frac14\right)\right)$$

$$=\frac\pi2+\tan^{-1}\left(\frac14\right)$$

Hence, the angle subtended by $$PQ$$ at the origin is $$\boxed{\frac\pi2+\tan^{-1}\left(\frac14\right)}$$