An ellipse having foci at (3, 1) and (1, 1) p | vedclass

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(A) Let the foci be $S(3, 1)$ and $S'(1, 1)$,and the point on the ellipse be $P(1, 3)$.The distance $PS = \sqrt{(3-1)^2 + (1-3)^2} = \sqrt{2^2 + (-2)^

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$\sqrt{2} - 1$

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(A) Let the foci be $S(3, 1)$ and $S'(1, 1)$,and the point on the ellipse be $P(1, 3)$.The distance $PS = \sqrt{(3-1)^2 + (1-3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = 2\sqrt{2}$.The distance $PS' = \sqrt{(1-1)^2 + (1-3)^2} = \sqrt{0^2 + (-2)^2} = 2$.By the definition of an ellipse,the sum of the distances from any point on the ellipse to the foci is constant and equal to the length of the major axis,$2a$.$2a = PS + PS' = 2\sqrt{2} + 2 \Rightarrow a = \sqrt{2} + 1$.The distance between the foci is $2ae = SS' = \sqrt{(3-1)^2 + (1-1)^2} = \sqrt{2^2 + 0^2} = 2$.Thus,$ae = 1$.Substituting $a = \sqrt{2} + 1$,we get $e = \frac{1}{\sqrt{2} + 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \sqrt{2} - 1$.

An ellipse having foci at $(3, 1)$ and $(1, 1)$ passes through the point $(1, 3)$. Then its eccentricity is

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