Let x^2=4ky, k0 be a parabola with vertex O( | vedclass

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(C) The parabola is $x^2=4ky$. The latus rectum $BC$ is the line $y=k$. The coordinates of $B$ and $C$ are $(-2k, k)$ and $(2k, k)$ respectively.Since

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$\frac{\sqrt{5}}{3}$

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(C) The parabola is $x^2=4ky$. The latus rectum $BC$ is the line $y=k$. The coordinates of $B$ and $C$ are $(-2k, k)$ and $(2k, k)$ respectively.Since $BD=DE=EC$ and $BC=4k$,we have $DE = \frac{4k}{3}$.The center of the ellipse $P$ is the midpoint of $DE$,which is $(0, k)$.The ellipse touches the parabola at the origin $O(0,0)$. Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.Since it passes through $O(0,0)$,we have $\frac{0}{a^2} + \frac{(-k)^2}{b^2} = 1$,so $b^2 = k^2$.The ellipse cuts $BC$ (the line $y=k$) at $D$ and $E$. Substituting $y=k$ into the ellipse equation gives $\frac{x^2}{a^2} + 0 = 1$,so $x = \pm a$.Thus,$D = (-a, k)$ and $E = (a, k)$. Since $DE = 2a = \frac{4k}{3}$,we have $a = \frac{2k}{3}$.Since $a Therefore,$e = \frac{\sqrt{5}}{3}$.

Let $x^2=4ky, k>0$ be a parabola with vertex $O(0,0)$. Let $BC$ be its latus rectum. An ellipse with center $P$ on $BC$ touches the parabola at $O$,and cuts $BC$ at points $D$ and $E$ such that $BD=DE=EC$ ($B, D, E, C$ in that order). The eccentricity of the ellipse is

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