Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.6 A Survey of Quadratic Surfaces - Exercises - Page 693: 41

Answer

$$\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1+\left(\frac{z}{\sqrt{27}}\right)^{2}.$$

Work Step by Step

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 $$\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1+\left(\frac{z}{\sqrt{27}}\right)^{2}.$$
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