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: 7

Answer

Foci:$\quad(\pm\sqrt{3},0)$ Eccentricity$:\qquad e =\displaystyle \frac{\sqrt{3}}{3}$ Directrices$:\quad x=\pm 3\sqrt{3}$ Graph:

Work Step by Step

$6x^{2}+9y^{2}=54\qquad /\div 54$ $\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{6}=1,\qquad$ We have a vertical major axis. $ a=6,b=\sqrt{6}$, Foci:$\quad F(\pm c,0)$ $c=\sqrt{a^{2}-b^{2}}=\sqrt{9-6}=\sqrt{3}$ Eccentricity$:\qquad e=\displaystyle \frac{c}{a}=\frac{\sqrt{3}}{3}$ Directrices$:\displaystyle \quad x=0\pm\frac{a}{e}$ $x=\displaystyle \pm\frac{3}{\frac{\sqrt{3}}{3}}$ $x=\pm 3\sqrt{3}$
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