College Algebra, Chapter 8, 8.4, Section 8.4, Problem 40

Find an equation for the ellipse that shares a vertex and a focus with the parabola $x^2 + y^2 = 100$ and has its other focus at the origin.

If we rewrite the parabola, we get $x^2 = -(y - 100)$. This parabola has a vertex of $(0, 100)$ and opens downward. Since $4p = 1$, we have $\displaystyle p = \frac{1}{4}$. It means that the focus is $\displaystyle \frac{1}{4}$ units below the vertex of the parabola. The reflecting focus will be $\displaystyle (0,100) \to \left(0, 100 - \frac{1}{4} \right) = \left( 0 , \frac{399}{4} \right)$.

Notice that if the other focus of the ellipse is at the origin, then the ellipse has a vertical major axis with $\displaystyle 2c = \frac{399}{4}$.

This shows that the center of the ellipse is $\displaystyle \frac{399}{8}$ units from the focus. So the center of the ellipse is at $\displaystyle \left( 0, \frac{399}{8} \right)$. Also, if one of the vertex is at $(0, 100)$ then the other vertex will be $50$ units from the center. This gives us $a = 50$. Then, the vertex is $\displaystyle \left( 0, \frac{399}{8} - 50 \right) \to \left( 0, \frac{-1}{8} \right)$.

To solve for $b$, we know that $c^2 = a^2 - b^2$


$
\begin{equation}
\begin{aligned}

b^2 =& a^2 - c^2
\\
\\
b^2 =& 50^2 - \left( \frac{399}{8} \right)^2
\\
\\
b^2 =& \frac{799}{64}
\\
\\
b =& \frac{\sqrt{799}}{8}

\end{aligned}
\end{equation}
$


Therefore, the equation of the ellipse is..


$
\begin{equation}
\begin{aligned}

\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} =& 1
\\
\\
\frac{(x - 0)^2}{\displaystyle \frac{799}{64}} + \frac{\displaystyle \left( y - \frac{399}{8} \right)^2}{50^2} =& 1
\\
\\
\frac{64x^2}{799} + \frac{\displaystyle \left( y - \frac{399}{8} \right)^2 }{2500} =& 1

\end{aligned}
\end{equation}
$