Answer
$r=\frac{ep}{1+e\cos\theta}$
Work Step by Step
We know that $d(F,P)=e\cdot d(D,P)$ and that $d(D,P)=p-r\cos\theta$.
Hence,
$r=e\cdot(p-r\cos\theta)\\r=ep-er\cos\theta\\r+er\cos\theta=ep\\r(1+e\cos\theta)=ep\\r=\frac{ep}{1+e\cos\theta}$
Precalculus (10th Edition)
by Sullivan, Michael
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: 43
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