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: 18

Answer

$\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ Foci: $\qquad (\pm 5,0)$ Directrices: $\quad x=\displaystyle \pm\frac{16}{5}$

Work Step by Step

Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$ Write in standard form and read a,b. $9x^{2}-16y^{2}=144\qquad /\div 144$ $\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1\qquad$ We have a horizontal axis, $a=4, b=3$ Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{16+9}=5$ Foci: $\quad (\pm c, 0)=\qquad (\pm 5,0)$ Vertices: $\quad (\pm a, 0)=\qquad (\pm 4, 0)$ Asymptotes: $\quad y=\displaystyle \pm\frac{b}{a}x \Rightarrow \quad y=\displaystyle \pm\frac{3}{4}x$ Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{5}{4}$ Directrices: $\quad x=0\displaystyle \pm\frac{a}{e}=\pm\frac{16}{5}$
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