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 685: 9



Work Step by Step

The eccentricity of the ellipse $\dfrac{x^2}{m^2}+\dfrac{y^2}{n^2}=1$ when $m \gt n$ is given by: $e=\dfrac{\sqrt {m^2-n^2}}{m}$ The foci of the ellipse are: $(\pm me,0)$ and the directrices are given as: $x=\pm \dfrac{m}{e}$ Given: $e=0.5$ ; foci: $(0,\pm 3)$ Now, $m=\dfrac{c}{e}=\dfrac{3}{0.5}=6$ and $q^2=m^2-c^2=36-9=27$ Here, the equation of the ellipse is: $\dfrac{x^2}{27}+\dfrac{y^2}{36}=1$
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