Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.6 Exercises - Page 708: 8

Answer

$$r = \frac {18}{1 - 3sin(\theta)}$$

Work Step by Step

1. Determine the equation that we are going to use: $$r = -6csc(\theta) = -6 \frac 1 {sin(\theta)} \longrightarrow rsin(\theta) = -6 \longrightarrow y = -6$$ Since the directrix is defined by $y = -6$, we are going to use $sin(\theta)$ in the equation. Since $-6$ is negative, the equation will have "$-sin(\theta)$": $$r = \frac{ed}{1 - esin(\theta)}$$ 2. Substitute the given values for $d$ (directrix) and e (eccentricity): (d = 6) $$r = \frac{3 (6)}{1 - 3sin(\theta)} = \frac {18}{1 - 3sin(\theta)}$$
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