Finding the Focus of a Parabolic Reflector

Energy hitting the surface of a parabolic reflector is concentrated at the focal point of the reflector (Figure 1). If the surface of a parabolic reflector is described by equation $\frac{x^2}{100}+\frac{y^2}{100}=\frac{z}{4}$, where is the focal point of the reflector?

Figure 1. Energy reflects off of the parabolic reflector and is collected at the focal point. (credit: modification of CGP Grey, Wikimedia Commons)

Solution

Since $z$ is the first-power variable, the axis of the reflector corresponds to the $z$-axis. The coefficients of $x^2$ and $y^2$ are equal, so the cross-section of the paraboloid perpendicular to the $z$-axis is a circle. We can consider a trace in the $xz$-plane or the $yz$-plane; the result is the same. Setting $y=0$, the trace is a parabola opening up along the $z$-axis, with standard equation $x^2=4pz$, where $p$ is the focal length of the parabola. In this case, this equation becomes $x^2=100\cdot\frac{z}{4}=4pz$ or $25=4p$. So $p$ is $6.25$ m, which tells us that the focus of the paraboloid is $6.25$ m up the axis from the vertex. Because the vertex of this surface is the origin, the focal point is $(0,0,6.25)$.