Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - Review - Exercises - Page 883: 38

Answer

$1=\dfrac{x^2}{4/3}+\dfrac{(y+5/3)^2}{16/9}+\dfrac{z^2} {4/3}$ It is an ellipsoid centered at $ ( 0, -\frac{5}{3}, 0)$

Work Step by Step

Use the distance formula from P to plane $y=1$ is twice distance from $P$ to point $( 0,-1,0)$ . $\sqrt {(x-x)^2+(y-1)^2+(z-z)^2}=2\sqrt {\sqrt {(x-0)^2+(y-(-1))^2+(z-0)^2}}$ $(y-1)^2=4 (x^2+4y^2+8y+4+4z^2$ $y^2-2y+1=4 (x^2+4y^2+8y+4+4z^2$ $-3=4x^2+3(y^2+10/3y+(10/6)^2)+4z^2$ $1=\dfrac{x^2}{4/3}+\dfrac{(y+5/3)^2}{16/9}+\dfrac{z^2} {4/3}$ It is an ellipsoid centered at $ ( 0, -\frac{5}{3}, 0)$
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