Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 32

Answer

$x^2=-12(y-3)$

Work Step by Step

Distance of the point $(x,y)$ from the line $y=6$ is $|y-6|$ Therefore, the equation of the parabola is $|y-6|=\sqrt {x^2+y^{2}}$ $x^2=-12y+36$ $x^2=-12(y-3)$
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