Let x^2=4 k y, k0 be a parabola with vert | vedclass

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Question

(c)

Given, $x^2=4 k y$

$B C$ is latusrectum.

$B C=4 k$

$B D=D E=E C$

$D E=\frac{B C}{3}=\frac{4 k}{3}$

$P$ is centre of ellipse.

Vedclass · Accepted Answer

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

Vedclass · Answer

(c)

Given, $x^2=4 k y$

$B C$ is latusrectum.

$B C=4 k$

$B D=D E=E C$

$D E=\frac{B C}{3}=\frac{4 k}{3}$

$P$ is centre of ellipse.

$P E=\frac{2 k}{3}$

$O P^2=k$

$\because$ Eccentricity of ellipse

$\sqrt{1-\frac{P E^2}{O P^2}}=\sqrt{1-\frac{4 k^2}{9 k^2}}$

$e=\sqrt{\frac{9-4}{9}}=\frac{\sqrt{5}}{3}$

Let $x^2=4 k y, k>0$ be a parabola with vertex $A$. Let $B C$ be its latusrectum. An ellipse with centre on $B C$ touches the parabola at $A$, and cuts $B C$ at points $D$ and $E$ such that $B D=D E=E C(B, D, E, C$ in that order). The eccentricity of the ellipse is

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