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: 34



Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.