Ellipse application problem. I am not sure where to begin with this problem. Is there an easier way of doing this problem?

Question

An express way is built in the shape of a semielliptical arch and is to have a span of one hundred feet. The height of the arch, at a distance of forty feet from center, is ten feet. Find the height of arch at the center.

Answer 1

Let $|AC|=L=100$, $|OA|=|OC|=\tfrac12L$, $|OE|=d=40$, $|BE|=h=10$.

Consider a circle $\mathcal C$ through the endpoints of $AC$ centered at $O$, with the radius $\tfrac12L$. The extension of the line $EB$ intersects with the circle $\mathcal C$ in $D$. Let $|BD|=g$, then $k=\frac h{h+g}$ is an $y$-scaling factor that transforms the circle $\mathcal C$ into the sought ellipse through $A,B,C$, and its height $H$ at the center is then $k\cdot\tfrac12L$.

So, first, we need to find $g$. By the Stewart’s theorem,

\begin{align} \triangle DOE:\quad |OE|^2\cdot|BD|+|OD|^2\cdot|BE| &=|DE|\cdot(|OB|^2+|BD|\cdot|BE|) \tag{1}\label{1} ,\\ \triangle BOE:\quad |OB|^2&=d^2+h^2 \tag{2}\label{2} , \end{align} and it follows that \begin{align} g&=\tfrac12\,\sqrt{L^2-4d^2}-h =20 \tag{3}\label{3} ,\\ k&=\tfrac13 \tag{4}\label{4} ,\\ H&=\tfrac{50}3 \tag{5}\label{5} . \end{align}