Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.1 - Three-Dimensional Coordinate Systems - Exercises 12.1 - Page 696: 64



Work Step by Step

Here, we have one point $P(0,0,2)$ and $xy$ -plane, this means that point: $(x,y,0)$. Formula to calculate the distance between two points is: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Now, $\sqrt{(x-0)^2+(y-0)^2+(z-2)^2}=\sqrt{(x-x)^2+(y-y)^2+(z-0)^2}$ $ \implies x^2+y^2+(z-2)^2=z^2$ so, $x^2+y^2-4z+4=0$ Thus, $z=\dfrac{x^2}{4}+\dfrac{y^2}{4}+1$
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