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(C) Let the equation of the ellipse be $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Since it passes through $(0,0)$,$(1,0)$,and $(0,2)$,we substit

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$\sqrt{2}-1$

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(C) Let the equation of the ellipse be $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Since it passes through $(0,0)$,$(1,0)$,and $(0,2)$,we substitute these points:$1$) $\frac{h^2}{a^2} + \frac{k^2}{b^2} = 1$$2$) $\frac{(1-h)^2}{a^2} + \frac{k^2}{b^2} = 1 \implies \frac{1-2h+h^2}{a^2} + \frac{k^2}{b^2} = 1$. Subtracting $(1)$ from $(2)$ gives $\frac{1-2h}{a^2} = 0 \implies h = \frac{1}{2}$.$3$) $\frac{h^2}{a^2} + \frac{(2-k)^2}{b^2} = 1 \implies \frac{h^2}{a^2} + \frac{4-4k+k^2}{b^2} = 1$. Subtracting $(1)$ from $(3)$ gives $\frac{4-4k}{b^2} = 0 \implies k = 1$.Thus,the center is $(\frac{1}{2}, 1)$. The equation is $\frac{(x-1/2)^2}{a^2} + \frac{(y-1)^2}{b^2} = 1$. Substituting $(0,0)$ gives $\frac{1/4}{a^2} + \frac{1}{b^2} = 1$.Since a focus lies on the $Y$-axis $(x=0)$,the distance from the center $(\frac{1}{2}, 1)$ to the focus $(0, 1)$ is $ae = \frac{1}{2}$.Thus $a^2e^2 = \frac{1}{4}$. Using $b^2 = a^2(1-e^2)$,we have $\frac{1}{4a^2} + \frac{1}{a^2(1-e^2)} = 1 \implies \frac{1-e^2+4}{4a^2(1-e^2)} = 1$. Using $a^2 = \frac{1}{4e^2}$,we get $\frac{5-e^2}{4(\frac{1}{4e^2})(1-e^2)} = 1 \implies \frac{e^2(5-e^2)}{1-e^2} = 1 \implies 5e^2 - e^4 = 1 - e^2 \implies e^4 - 6e^2 + 1 = 0$.Solving for $e^2$: $e^2 = \frac{6 \pm \sqrt{36-4}}{2} = 3 \pm \sqrt{8} = 3 \pm 2\sqrt{2}$.Since $e < 1$,$e^2 = 3 - 2\sqrt{2} = (\sqrt{2}-1)^2$. Thus $e = \sqrt{2}-1$.

An ellipse with its minor and major axes parallel to the coordinate axes passes through $(0,0)$,$(1,0)$,and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is

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