Activity 6.13 Slicing the Hyperboloids.
Use the same slicing method as in the previous subsections to study the two surfaces
\begin{equation*}
\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 1
\qquad \text{and} \qquad
\frac{z^2}{16} - \frac{x^2}{4} - \frac{y^2}{9} = 1,
\end{equation*}
both of which come from equation (6.1). Discover why the first is called a hyperboloid of one sheet while the second is called a hyperboloid of two sheets. Pay particular attention to the traces in the horizontal planes \(z = c\text{:}\) they are what tells the two surfaces apart.
5
For the first surface \(A = 1/4\text{,}\) \(B = 1/9\text{,}\) \(C = -1/16\text{,}\) \(D = 0\text{,}\) and \(E = 1\text{;}\) for the second surface \(A = -1/4\text{,}\) \(B = -1/9\text{,}\) \(C = 1/16\text{,}\) \(D = 0\text{,}\) and \(E = 1\text{.}\) For the ellipsoid all three squared terms were positive; here exactly one squared term is negative for the first surface and exactly two are negative for the second. Keep this count in mind as you work through the traces.
(a)
For the surface \(\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 1\text{,}\) find the trace in the plane \(z = c\) for an arbitrary constant \(c\text{,}\) and the traces in the coordinate planes \(x = 0\) and \(y = 0\text{.}\) For which values of \(c\) does the plane \(z = c\) actually intersect the surface? Identify each trace as an ellipse, parabola, or hyperbola.
(b)
Repeat the analysis for the surface \(\frac{z^2}{16} - \frac{x^2}{4} - \frac{y^2}{9} = 1\text{.}\) For which values of \(c\) does the plane \(z = c\) intersect the surface this time? What happens for \(c\) between those values?
(c)
Using only the traces in the planes \(z = c\text{,}\) explain why the first surface is called a hyperboloid of one sheet and the second a hyperboloid of two sheets. Can you predict the number of sheets directly from the signs in the equation?
