Let the line y - x = l intersect the ellipse $$\f\frac{x^{2}}{2}+\f\frac{y^{2}}{1}=$$ at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is:

We need to find the angle subtended by chord AB at the center of the ellipse $$\f\frac{x^2}{2} + y^2 = 1$$, where the line $$y - x = 1$$ (that is, $$y = x + 1$$) meets the ellipse at points A and B.

First, substituting $$y = x + 1$$ into the ellipse equation gives

$$ \f\frac{x^2}{2} + (x+1)^2 = 1 $$

which simplifies to

$$ \f\frac{x^2}{2} + x^2 + 2x + 1 = 1 \quad\Longrightarrow\quad \f\frac{3x^2}{2} + 2x = 0 \quad\Longrightarrow\quad x(3x + 4) = 0,$$

so that $$x = 0$$ or $$x = -\f\frac{4}{3}\,.$$

When $$x = 0$$, we have $$y = 1$$, hence $$A = (0,1)$$. When $$x = -\f\frac{4}{3}$$, we get $$y = -\f\frac{4}{3} + 1 = -\f\frac{1}{3}$$, so $$B = \bigl(-\tf\frac{4}{3}, -\tf\frac{1}{3}\bigr)\,.$

Next, we denote $$\overrightarrow{OA} = (0,1)$$ and $$\overrightarrow{OB} = \bigl(-$$\tfrac{4}{3}$$, -$$\tfrac{1}{3}$$\bigr)\,$$ and use the dot‐product formula

$$ $$\cos$$$$\theta = \frac{\overrightarrow{OA}$$$$\cdot$$\overrightarrow{OB}}{\lvert\overrightarrow{OA}\rvert\,\lvert\overrightarrow{OB}\rvert}\,. $$

Since $$\overrightarrow{OA}$$\cdot$$\overrightarrow{OB} = 0$$\cdot$$\bigl(-$$\tfrac{4}{3}$$\bigr) + 1$$\cdot$$\bigl(-$$\tfrac{1}{3}$$\bigr) = -$$\frac{1}{3}$$,$$

$$\lvert\overrightarrow{OA}\rvert = 1,\qquad \lvert\overrightarrow{OB}\rvert = $$\sqrt{\tfrac{16}{9}+\tfrac{1}{9}$$} = $$\frac{\sqrt{17}$$}{3},$$

we obtain

$$ $$\cos$$$$\theta = \frac{-\tfrac{1}{3}$$}{1$$\cdot$$($$\sqrt{17}$$/3)} = -$$\frac{1}{\sqrt{17}$$}\,. $$

Because $$$$\cos$$$$\theta$$ = -$$\frac{1}{\sqrt{17}$$}\,, $$ it follows that

$$ $$\theta = \pi - \cos^{-1}\!\bigl(\tfrac{1}{\sqrt{17}}\bigr$$)\,. $$

On the other hand, if we set $$$$\tan$$$$\phi$$ = 4$$ with $$$$\phi = \tan^{-1}(4$$)\,, $$ then in a right triangle with opposite side 4 and adjacent side 1 the hypotenuse is $$$$\sqrt{17}$$$$, so $$$$\cos$$$$\phi = \tfrac{1}{\sqrt{17}$$}$$ and hence $$$$\cos^{-1}\!\bigl(\tfrac{1}{\sqrt{17}}\bigr$$) = $$\tan^{-1}(4$$)\,.$$

Moreover, by the complementary‐angle identity, $$$$\tan^{-1}(4$$) = $$\tfrac{\pi}{2} - \tan^{-1}\!\bigl(\tfrac{1}{4}\bigr$$)\,. $$ Therefore

$$ $$\theta = \pi$$ - \Bigl($$\tfrac{\pi}{2} - \tan^{-1}\!\bigl(\tfrac{1}{4}\bigr$$)\Bigr) = $$\frac{\pi}{2} + \tan^{-1}\!\bigl(\tfrac{1}{4}\bigr$$)\,. $$

The correct answer is Option 4: $$$$\frac{\pi}{2} + \tan^{-1}\!\Bigl(\frac{1}{4}\Bigr$$)\,. $$