Answer
$\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
Work Step by Step
Given: $x^2+y=100$
Re-arrange as: $x^2=4(\dfrac{-1}{4})(y-100)$
Here, we have the vertex: $(0,100)$ and the focus is: $(0,100-\dfrac{1}{4})=(0,\dfrac{399}{4})$
Since, the mid-point of the foci is the center of the ellipse, we get
Center: $(0,\dfrac{399}{8})$
Thus, the equation of the ellipse is: $\dfrac{(x-0)^2}{a^2}+\dfrac{(y-\dfrac{399}{8})^2}{b^2}=1$
or, $\dfrac{x^2}{b^2}+\dfrac{(y-\dfrac{399}{8})^2}{a^2}=1$ ...(1)
The eccentricity of the ellipse is: $ae=\dfrac{399}{8}$
Thus, we have $a=100-\dfrac{399}{8}=\dfrac{401}{8}$
and
$b^2=a^2(1-e^2)=a^2-(ae)^2$
This gives: $b^2=(\dfrac{401}{8})^2-(\dfrac{399}{8})^2=25$
Therefore, equation (1) of the ellipse becomes:
$\dfrac{x^2}{25}+\dfrac{(y-\dfrac{399}{8})^2}{(\dfrac{401}{8})^2}=1$
Hence, $\dfrac{x^2}{25}+\dfrac{(8y-399)^2}{160,801}=1$
