Ellipse

Concept:The angle subtended by a chord at the centre is the angle between the radius vectors to its endpoints.Solution:The line is $y-x=1$ i.e. $y=x+1$. Substitute in ellipse $\frac{x^2}{2}+y^2=1$:$\frac{x^2}{2}+(x+1)^2=1$. Multiply by 2: $x^2+2(x+1)^2=2$. Expand: $x^2+2(x^2+2x+1)=2 \Rightarrow x^2+2x^2+4x+2=2 \Rightarrow 3x^2+4x=0$. Factor: $x(3x+4)=0 \Rightarrow x=0$ or $x=-\frac{4}{3}$. Corresponding $y$: for $x=0$, $y=1$; for $x=-\frac{4}{3}$, $y=-\frac{1}{3}$. Thus points $A(0,1)$ and $B\left(-\frac{4}{3}, -\frac{1}{3}\right)$. Center of ellipse is $O(0,0)$. Vector $\overrightarrow{OA}$ makes angle $90^\circ$ (positive y‑axis). Vector $\overrightarrow{OB}$ has angle $\theta$ from positive x‑axis: $\tan\theta = \frac{-1/3}{-4/3}=\frac{1}{4}$, and since both coordinates negative, $\theta = \pi + \tan^{-1}\left(\frac{1}{4}\right)$. Angle $AOB = \theta - 90^\circ = \left(\pi + \tan^{-1}\frac{1}{4}\right) - \frac{\pi}{2} = \frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$. Answer:$\frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)$ (Option B)