Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Review - Exercises - Page 843: 35

Answer

Ellipsoid

Work Step by Step

Here, we have: $4x^2+4y^2-8y+z^2=0$ Rewrite as: $4x^2+4(y^{2}-2y) + z^{2} =0$ or, $4x^2 +4(y^2-2y+1) +z^2=4$ or, $x^2+(y-1)^2 + \dfrac{z^2}{4}=1$ On comparing the above form with $ \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^2}{c^2}=1$ we find that we have an ellipsoid.
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