Answer
$\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
Work Step by Step
Re-arrange the given equation as: $x^2=4(\dfrac{-1}{4})(y-100)$
We are given that Vertex: $(0,100)$ and Focus: $F(0,100-\dfrac{1}{4})=F(0,\dfrac{399}{4})$
As the mid-point of the foci is center of the ellipse, then we have:
Center: $(0,\dfrac{399}{8})$
Now, the form of the ellipse as follows:
$\dfrac{(x-0)^2}{a^2}+\dfrac{(y-\dfrac{399}{8})^2}{b^2}=1$
or, $\dfrac{x^2}{b^2}+\dfrac{(y-\dfrac{399}{8})^2}{a^2}=1$
The eccentricity of the ellipse is defined as:
$ae=\dfrac{399}{8}$
$\implies a=100-\dfrac{399}{8}=\dfrac{401}{8}$
and $b^2=a^2(1-e^2)=a^2-(ae)^2$
$\implies b^2=(\dfrac{401}{8})^2-(\dfrac{399}{8})^2=25$
So, we get the equation of ellipse as:
$\dfrac{x^2}{25}+\dfrac{(y-\dfrac{399}{8})^2}{(\dfrac{401}{8})^2}=1$
Hence, $\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
