Answer
$\frac{x^2}{25}+\frac{(y-49.875)^2}{(50.125)^2}=1$
Work Step by Step
Step 1. Find the vertex and focus of the parabola: $x^2+y=100$ or $x^2=-(y-100)$, we have $4p=-1$ and $p=-\frac{1}{4}$ and it has a vertex at $(0,100)$, focus at $(0, 100-\frac{1}{4})$
Step 2. With the condition that another focus is at the origin $(0,0)$, we get the center at $(0, 50-\frac{1}{8})$
Step 3. The distance between the foci is $2c=100-\frac{1}{4}$, thus $c=50-\frac{1}{8}$
Step 4. The distance between the center and the vertex is $a=100-(50-\frac{1}{8})=50+\frac{1}{8}$
Step 5. The shorter axis $b^2=a^2-c^2=(50+\frac{1}{8})^2-(50-\frac{1}{8})^2=25$
Step 6. With the results for the center and two axis lengths, we can write the equation of the ellipse as
$\frac{x^2}{25}+\frac{(y-50+\frac{1}{8})^2}{(50+\frac{1}{8})^2}=1$
or
$\frac{x^2}{25}+\frac{(y-49.875)^2}{(50.125)^2}=1$
