Answer
$x^2-3y^2+12y-9=0$,
Hyperbola.
Work Step by Step
$r=\frac{12}{4+8sin\theta}$,
$r=\frac{3}{1+2sin\theta}$,
$r+2r\ sin\theta=3$,
$r=3-2r\ sin\theta$,
$r^2=(3-2r\ sin\theta)^2$,
$x^2+y^2=(3-2y)^2$,
$x^2-3y^2+12y-9=0$,
Hyperbola.
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 - Section 9.6 Polar Equations of Conics - 9.6 Assess Your Understanding - Page 704: 30
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