Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.7 - Conics in Polar Coordinates - Exercises 11.7 - Page 686: 43

Answer

Graph: .

Work Step by Step

Sine in the denominator $\Rightarrow\qquad$ the main axis is vertical. We want $1\pm...$ in the denominator, $\displaystyle \frac{8\div 2}{(2-2\sin\theta)\div 2}=\frac{4}{1-\sin\theta}=\frac{-4(-1)}{1+(-1)\sin\theta}$ Compare with $r=\displaystyle \frac{ke}{1\pm e\sin\theta}$ $e=1, \qquad$ so this is a parabola. $k=4$ $y=-4$ is the directrix, so the parabola opens up. The vertex is halfway between the directrix and focus, 2 units below the focus at the origin, on the y-axis $(\theta=3\pi/2)$, with polar coordinates: $(2,3\pi/2).$
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