## Calculus (3rd Edition)

$$\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1+\left(\frac{z}{\sqrt{27}}\right)^{2}.$$
By observing Figure 16(A), we get that $$a=4, \quad b=6.$$ Plugging into the equation on page 691, we get the hyperboloid $$\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1+\left(\frac{z}{c}\right)^{2}.$$ To find $c$, we use the fact that the surface passes through the point $(0,12,9)$. By substitution, we obtain that $c=\sqrt{27}$, and hence the equation is given by $$\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1+\left(\frac{z}{\sqrt{27}}\right)^{2}.$$