Calculus (3rd Edition)

The equation of a hyperboloid is $\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=\left(\frac{z}{c}\right)^{2}+1$ and to find the trace we freeze one of the three variables, for example $z=z_0$. Then the equation of the hyperboloid becomes $$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=\left(\frac{z_0}{c}\right)^{2}+1$$ which represent an ellipse. So the statement is false.