## Calculus (3rd Edition)

The equation $$x^2-3y^2-9z^2=1$$ can be rewritten in the form $$x^2- \frac{y^2}{1/3}-\frac{z^2}{1/9}=1\\ \frac{y^2}{1/3}+\frac{z^2}{1/9}=x^2-1 .$$ Hence the equation is a hyperboloid of two sheets (see equations on page 688).