Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 10 - Analytic Geometry - 10.6 Polar Equations of Conics - 10.6 Assess Your Understanding - Page 685: 31



Work Step by Step

After multiplying with the denominator of the fraction: $r(2-\sin{\theta})=8\\2r-r\sin{\theta}=8\\2r=8+r\sin{\theta}\\4r^2=(8+r\sin{\theta})^2$. We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$. Hence $4r^2=(8+r\sin{\theta})^2$ becomes: $4(x^2+y^2)=(8+y)^2\\4x^2+4y^2=64+16y+y^2\\4x^2+3y^2-64-16y=0$
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