University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.6 - Cylinders and Quadric Surfaces - Exercises - Page 636: 25

Answer

Elliptical (circular) cone. See image: .

Work Step by Step

Comparing to Table 11.1, the equation is of the form $\displaystyle \frac{x^{2}}{1^{2}}+\frac{y^{2}}{1^{2}}=\frac{z^{2}}{1}$ this is an elliptical cone, in which the traces in planes parallel to the xy plane ( $z=k$) are circles of radius $k$.
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