Let the line y-x=1 intersect the ellipse frac | vedclass

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(B) The equation of the line is $y = x + 1$. Substituting this into the ellipse equation $\frac{x^2}{2} + y^2 = 1$:$\frac{x^2}{2} + (x+1)^2 = 1$$\frac

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$\frac{\pi}{2}+	an^{-1}(\frac{1}{4})$

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(B) The equation of the line is $y = x + 1$. Substituting this into the ellipse equation $\frac{x^2}{2} + y^2 = 1$:$\frac{x^2}{2} + (x+1)^2 = 1$$\frac{x^2}{2} + x^2 + 2x + 1 = 1$$\frac{3x^2}{2} + 2x = 0$$x(\frac{3x}{2} + 2) = 0$So,$x = 0$ or $x = -\frac{4}{3}$.If $x = 0$,then $y = 1$,so $A = (0, 1)$.If $x = -\frac{4}{3}$,then $y = -\frac{4}{3} + 1 = -\frac{1}{3}$,so $B = (-\frac{4}{3}, -\frac{1}{3})$.The center of the ellipse is $O(0, 0)$.The slope of $OA$ is $m_1 = \frac{1-0}{0-0} = \infty$ (vertical line,angle is $\frac{\pi}{2}$).The slope of $OB$ is $m_2 = \frac{-1/3 - 0}{-4/3 - 0} = \frac{1}{4}$.The angle $	heta$ that $OB$ makes with the negative $x$-axis is $	an^{-1}(\frac{1}{4})$.The angle $\angle AOB$ is the angle between the $y$-axis and the line $OB$,which is $\frac{\pi}{2} + 	an^{-1}(\frac{1}{4})$.

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

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