Question

Let $x^2=4ky,k>0$, be a parabola with vertex $A$. Let $BC$ be its latus rectum. An ellipse with center on $BC$ touches the parabola at $A$, 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

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{\sqrt{5}}{3}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is C $\frac{\sqrt{5}}{3}$

Given :
The equation of parabola is $x^2=4ky$


$BD=DE=EC$
So,
Vertex $=(0,0)$ and $a=k$
Latus rectum $=4a=4k$
We know,
$BD+DE+EC=4k\Rightarrow DE=\frac{4k}{3}$
minor axis $=\frac{4k}{3}$
major axis $=2a=2k$

So, the eccentricity, e $=\sqrt{1-\frac{(minoraxis{)}^2}{(majoraxis{)}^2}}={\sqrt{1-\left[\frac{\left(\frac{4k}{3}\right)}{2k}\right]}}^2$
$e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$