## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 54

#### Answer

$$r= -\frac{12}{2+ 3\cos \theta}.$$

#### Work Step by Step

The polar equation of the conic of eccentricity $e \gt0$, with focus at the origin and directrix $x = d$ is $$r=\frac{e d}{1+e \cos \theta}.$$ Now, since $e=3/2$ and $x=-4$, then the polar equation is $$r=\frac{-6}{1+ (3/2)\cos \theta} =-\frac{12}{2+ 3\cos \theta}.$$

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