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(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where the center is at $(0, b)$ relative to the diameter. Since it touc

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$\sqrt{\frac{2}{3}}$

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(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where the center is at $(0, b)$ relative to the diameter. Since it touches the diameter $y=0$,the ellipse is $\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} = 1$. The semi-circle is $x^2 + y^2 = r^2$ for $y \ge 0$. Substituting $x^2 = a^2(1 - \frac{(y-b)^2}{b^2})$ into the circle equation: $a^2(1 - \frac{(y-b)^2}{b^2}) + y^2 = r^2$ $a^2 - \frac{a^2}{b^2}(y^2 - 2by + b^2) + y^2 = r^2$ $(1 - \frac{a^2}{b^2})y^2 + \frac{2a^2}{b}y + (a^2 - a^2 - r^2) = 0$ $(1 - \frac{a^2}{b^2})y^2 + \frac{2a^2}{b}y - r^2 = 0$. For tangency,the discriminant $D = 0$: $(\frac{2a^2}{b})^2 - 4(1 - \frac{a^2}{b^2})(-r^2) = 0$ $\frac{4a^4}{b^2} + 4r^2 - \frac{4a^2r^2}{b^2} = 0$ $a^4 + b^2r^2 - a^2r^2 = 0 \Rightarrow b^2 = \frac{a^2r^2 - a^4}{r^2} = a^2(1 - \frac{a^2}{r^2})$. Area $A = \pi ab = \pi a^2 \sqrt{1 - \frac{a^2}{r^2}}$. Let $u = a^2$,then $A^2 = \pi^2 u^2(1 - \frac{u}{r^2}) = \pi^2(u^2 - \frac{u^3}{r^2})$. For maximum area,$\frac{d(A^2)}{du} = 2u - \frac{3u^2}{r^2} = 0 \Rightarrow u = \frac{2r^2}{3}$. So $a^2 = \frac{2r^2}{3}$ and $b^2 = \frac{2r^2}{3}(1 - \frac{2}{3}) = \frac{2r^2}{9}$. Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2r^2/9}{2r^2/3}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}$.

An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area,its eccentricity is

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