Answer
$\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
Work Step by Step
We need to re-arrange the given equations as: $x^2=4(\dfrac{-1}{4})(y-100)$
Vertex: $(0,100)$ and Focus:F $(0,100-\dfrac{1}{4})=F(0,\dfrac{399}{4})$
with Centre: $(0,\dfrac{399}{8})$
The equation of the ellipse:
$\dfrac{x^2}{b^2}+\dfrac{(y-\dfrac{399}{8})^2}{a^2}=1$
Eccentricity $e$ of the ellipse: $ae=\dfrac{399}{8}$
$\implies a=100-\dfrac{399}{8}=\dfrac{401}{8}$;
Also, we have $b^2=a^2(1-e^2)=a^2-(ae)^2$
This gives: $b^2=(\dfrac{401}{8})^2-(\dfrac{399}{8})^2=25$
Now, the equation of the ellipse is:
$\dfrac{x^2}{25}+\dfrac{(y-399/8)^2}{(401/8)^2}=1$
Hence, we have
$\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
