Question 10.6.6: Sketching a Hyperboloid of One Sheet Draw a graph of the qua...
Sketching a Hyperboloid of One Sheet
Draw a graph of the quadric surface
\frac{x^2}{4}+y^2-\frac{z^2}{2}=1 .The traces in the coordinate plane are as follows:
y z \text {-plane }(x=0): y^2-\frac{z^2}{2}=1 \text { (hyperbola) }(see Figure 10.59a),
x y \text {-plane }(z=0): \frac{x^2}{4}+y^2=1 \text { (ellipse) }and x z \text {-plane }(y=0): \frac{x^2}{4}-\frac{z^2}{2}=1 \text { (hyperbola). }
Further, notice that the trace of the surface in each plane z=k \text { (parallel to the } x y \text {-plane) }
is also an ellipse:
\frac{x^2}{4}+y^2=\frac{k^2}{2}+1.
Finally, observe that the larger k is, the larger the axes of the ellipses are. Adding this information to Figure 10.59a, we draw the surface seen in Figure 10.59b. We call this surface a hyperboloid of one sheet.
To plot this with a CAS, you could graph the two functions z=\sqrt{2\left(\frac{x^2}{4}+y^2-1\right)}
and z=-\sqrt{2\left(\frac{x^2}{4}+y^2-1\right)}. (See Figure 10.59c, where we have restricted the \text { z-range } to -10 \leq z \leq 10, to show the elliptical cross sections.) Notice that this plot looks more like a cone than the hyperboloid in Figure 10.59b. However, if you have drawn Figure 10.59b yourself, this plotting deficiency won’t fool you.
Alternatively, the parametric plot seen in Figure 10.59d, with x=2 \cos s \cosh t, y=\sin s \cosh t \text { and } z=\sqrt{2} \sinh t, \text { with } 0 \leq s \leq 2 \pi \text { and }-5 \leq t \leq 5 \text {, } shows the full hyperboloid with its elliptical and hyperbolic traces.
