Start by sketching the traces. To find the trace in the [latex]xy[/latex]-plane, set [latex]z=0[/latex]: [latex]\frac{x^2}{2^2}+\frac{y^2}{3^2}=1[/latex] (see Figure 1). To find the other traces, first set [latex]y=0[/latex] and then set [latex]x=0[/latex].

Figure 1. (a) This graph represents the trace of equation [latex]\frac{x^2}{2^2}+\frac{y^2}{3^2}=1[/latex] in the [latex]xy[/latex]-plane, when we set [latex]z=0[/latex]. (b) When we set [latex]y=0[/latex], we get the trace of the ellipsoid in the [latex]xz[/latex]-plane, which is an ellipse. (c) When we set [latex]x=0[/latex], we get the trace of the ellipsoid in the [latex]yz[/latex]-plane, which is also an ellipse.

Now that we know what traces of this solid look like, we can sketch the surface in three dimensions (Figure 2).

Figure 2. (a) The traces provide a framework for the surface. (b) The center of this ellipsoid is the origin.

To find the trace in the [latex]xy[/latex]-plane, set [latex]z=0[/latex]: [latex]x^2+\frac{y^2}{2^2}=0[/latex]. The trace in the plane [latex]z=0[/latex] is simply one point, the origin. Since a single point does not tell us what the shape is, we can move up the [latex]z[/latex]-axis to an arbitrary plane to find the shape of other traces of the figure.

The trace in plane [latex]z=5[/latex] is the graph of equation [latex]x^2+\frac{y^2}{2^2}=1[/latex], which is an ellipse. In the [latex]xz[/latex]-plane, the equation becomes [latex]z=5x^{2}[/latex]. The trace is a parabola in this plane and in any plane with the equation [latex]y=b[/latex].

In planes parallel to the [latex]yz[/latex]-plane, the traces are also parabolas, as we can see in the following figure.

Figure 4. (a) The paraboloid [latex]x^2+\frac{y^2}{2^2}=\frac{z}{5}[/latex]. (b) The trace in plane [latex]z=5[/latex]. (c) The trace in the [latex]xz[/latex]-plane. (d) The trace in the [latex]yz[/latex]-plane.

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

1. The [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex] terms are all squared, and are all positive, so this is probably an ellipsoid. However, let’s put the equation into the standard form for an ellipsoid just to be sure. We have[latex]16x^{2}+9y^{2}+16z^{2}=144[/latex].
   
   Dividing through by [latex]144[/latex] gives
   
   [latex]\frac{x^2}9+\frac{y^2}{16}+\frac{z^2}9=1[/latex].
   So, this is, in fact, an ellipsoid, centered at the origin.
2. We first notice that the [latex]z[/latex] term is raised only to the first power, so this is either an elliptic paraboloid or a hyperbolic paraboloid. We also note there are [latex]x[/latex] terms and [latex]y[/latex] terms that are not squared, so this quadric surface is not centered at the origin. We need to complete the square to put this equation in one of the standard forms. We have
   
   [latex]\begin{aligned} 9x^2-18x+4y^2+16y-36z+25&= 0\\ 9x^2-18x+4y^2+16y+25&=36z \\ 9(x^2-2x)+4(y^2+4y)+25&=36z \\ 9(x^2-2x+1-1)+4(y^2+4y+4-4)+25&=36z \\ 9(x-1)^2-9+4(y+2)^2-16+25&=36z \\ 9(x-1)^2+4(y+2)^2&=36z \\ \frac{(x-1)^2}4+\frac{(y+2)^2}9&=z. \end{aligned}[/latex]
   This is an elliptic paraboloid centered at [latex](1, 2, 0)[/latex].