Answer
$y^2-8x-16=0$.
Work Step by Step
$r=\frac{4}{1-cos\theta}$,
$r-r\ cos\theta=4$,
$r=4+r\ cos\theta$,
$r^2=(4+r\ cos\theta)^2$,
$x^2+y^2=(4+x)^2$,
$y^2-8x-16=0$.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 9 - Analytic Geometry - Cumulative Review - Review Exercises - Page 719: 30
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