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: 5

Answer

Foci:$\quad(0,\pm 1)$ Eccentricity$:\qquad e =\displaystyle \frac{1}{\sqrt{3}}$ Directrices$:\quad y=\pm 3$ Graph:

Work Step by Step

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