University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.6 - Conics in Polar Coordinates - Exercises - Page 591: 13



Work Step by Step

The eccentricity of the ellipse $\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=1$ when $p \gt q$ is given by: $e=\dfrac{\sqrt {p^2-q^2}}{p}$ The foci of the ellipse are: $(\pm pe,0)$ and the directrices are given as: $x=\pm \dfrac{p}{e}$ We are given that $e=0.24$ ; vertices: $(\sqrt 5,0)$ The directrix is given as: $x=\dfrac{9}{\sqrt 5}$ Now, $e=\dfrac{\sqrt 5}{3}$ Thus, the equation of the ellipse becomes: $(x-\sqrt 5)^2+y^2=\dfrac{5}{9} (x-\dfrac{9}{\sqrt 5})^2 \implies \dfrac{x^2}{9}+\dfrac{y^2}{4}=1$
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