A satellite dish has a parabolic cross-section that is \(4\) feet wide and \(1\) foot deep at its center. How far from the vertex should the receiver be mounted?
Solution.
Put the vertex at the origin with the dish opening upward, so the cross-section is \(x^2 = 4py\text{.}\) The dish is \(4\) feet wide and \(1\) foot deep, so the rim passes through the point \((2,1)\text{.}\) Substituting,
\begin{equation*}
2^2 = 4p(1) \;\Longrightarrow\; 4 = 4p \;\Longrightarrow\; p = 1.
\end{equation*}
The receiver belongs at the focus \((0,p) = (0,1)\text{,}\) that is, \(1\) foot above the vertex—which here happens to be exactly level with the rim.
