Let S(1,0) and S^{prime}(0,1) be the foci of | vedclass

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(D) The definition of an ellipse states that $SP + S^{\prime}P = 2a$. Given $SP + S^{\prime}P = 2$,we have $2a = 2$,so $a = 1$.The distance between th

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(D) The definition of an ellipse states that $SP + S^{\prime}P = 2a$. Given $SP + S^{\prime}P = 2$,we have $2a = 2$,so $a = 1$.The distance between the foci $SS^{\prime}$ is given by $\sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$.Since $SS^{\prime} = 2ae$,we have $2ae = \sqrt{2}$,which implies $e = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.The center of the ellipse is the midpoint of the foci $S$ and $S^{\prime}$,which is $(\frac{1+0}{2}, \frac{0+1}{2}) = (\frac{1}{2}, \frac{1}{2})$.The major axis lies along the line passing through $S(1,0)$ and $S^{\prime}(0,1)$. The slope of this line is $m = \frac{1-0}{0-1} = -1$.The unit vector along the major axis is $\vec{u} = \frac{S-S^{\prime}}{|S-S^{\prime}|} = \frac{(1, -1)}{\sqrt{2}} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.The endpoints of the major axis are at a distance $a=1$ from the center $C(\frac{1}{2}, \frac{1}{2})$ along the direction of the major axis.Thus,$A, A^{\prime} = C \pm a\vec{u} = (\frac{1}{2} \pm \frac{1}{\sqrt{2}}, \frac{1}{2} \mp \frac{1}{\sqrt{2}})$.Therefore,$x_1 = \frac{1}{2} + \frac{1}{\sqrt{2}}$ and $x_2 = \frac{1}{2} - \frac{1}{\sqrt{2}}$.Summing these,$x_1 + x_2 = (\frac{1}{2} + \frac{1}{\sqrt{2}}) + (\frac{1}{2} - \frac{1}{\sqrt{2}}) = 1$.

Let $S(1,0)$ and $S^{\prime}(0,1)$ be the foci of an ellipse such that $SP+S^{\prime} P=2$ for any point $P$ on the ellipse. If $A(x_1, y_1)$ and $A^{\prime}(x_2, y_2)$ are the end points of the major axis of this ellipse,then $x_1+x_2=$

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