so. hmmm we know the arc entry has a vertical clearance of 66 feet, and is 46feet wide
notice the picture below.. if we move the arc to the origin, we end up with a parabola like so, from -23, 0 to 23, 0
now [tex]\bf \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-----------------------------\\\\
y=a(x-0)^2+0\implies y=ax^2
\\\\\\
\textit{what's the value of "a"? well, we know }
\begin{cases}
x=23\\
y=-66
\end{cases}
\\\\\\
-66=a(23)^2\implies \cfrac{-6}{529}=a\qquad thus\qquad \boxed{y=-\cfrac{6}{529}x^2}[/tex]
