Precalculus (10th Edition)

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

Chapter 10 - Analytic Geometry - Chapter Review - Review Exercises - Page 700: 28

Answer

Ellipse See graph

Work Step by Step

We are given the polar equation: $r=\dfrac{6}{2-\sin\theta}$ Rewrite the equation: $r=\dfrac{\dfrac{6}{2}}{\dfrac{2-\sin\theta}{2}}$ $r=\dfrac{3}{1-\dfrac{1}{2}\sin\theta}$ The equation is in the form: $r=\dfrac{ep}{1-e\sin\theta}$ Determine $e$ using the denominator: $e=\dfrac{1}{2}$ Determine $p$ using the numerator: $ep=3$ $\dfrac{1}{2}p=3$ $p=6$ Because $e<1$, the conic is an ellipse. Its directrix is parallel to the polar axis at a distance of 6 units below the pole. The major axis is perpendicular to the directrix. Graph the curve:
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