2.6 Quadric surfaces  (Page 3/15)

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

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

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

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 $x^2=100·\frac{z}{4}=4pz$ 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 $\left(0,0,6.25\right).$