Answer
Cone
Work Step by Step
$x^2-y^2-z^2-4x-2z+3=0$
$x^2-4x-y^2-z^2-2z=-3$
$x^2-4x+4-y^2-z^2-2z-1=-3+4-1$
$(x-2)^2-y^2-(z+1)^2=0$
$(x-2)^2=y^2+(z+1)^2$
This is the equation of a cone along the x-axis with the vertex at (2,0,-1).
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 12 - Section 12.6 - Cylinders and Quadric Surfaces - 12.6 Exercises - Page 840: 36
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