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: 8

Answer

Foci: $(0,\pm 12)$ Eccentricity$:\qquad e =\displaystyle \frac{12}{13}$ Directrices$:\displaystyle \quad y=\pm\frac{169}{12}$ Graph:

Work Step by Step

$169x^{2}+25y^{2}=4225\qquad /\div 4225$ $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{169}=1,\qquad $ We have a vertical major axis. $a=5,b=13$, Foci: $(0,\pm c)$ $c=\sqrt{a^{2}-b^{2}}=\sqrt{169-25}=12$ Eccentricity$:\qquad e=\displaystyle \frac{c}{a}=\frac{12}{13}$ Directrices$:\quad y=0\displaystyle \pm\frac{a}{e}$ $y=\displaystyle \pm\frac{13}{(\frac{12}{13})}$ $y=\displaystyle \pm\frac{169}{12}$ $(y\approx 14.08)$
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