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: 19

Answer

$\displaystyle \frac{y^{2}}{8}-\frac{x^{2}}{8}=1$ Foci: $\qquad (0, \pm 4)$ Directrices: $\quad y=\pm 2$

Work Step by Step

Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$ Write in standard form and read a,b. $y^{2}-x^{2}=8\qquad /\div 8$ $\displaystyle \frac{y^{2}}{8}-\frac{x^{2}}{8}=1\qquad $ We have a vertical axis, $a=2\sqrt{2}, b=2\sqrt{2}$ Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{8+8}=4$ Foci: $\quad (0, \pm c)=\qquad (0, \pm 4)$ Vertices: $\quad (0, \pm a)=\qquad (0, \pm 2\sqrt{2})$ Asymptotes: $\quad y=\displaystyle \pm\frac{a}{b}x\Rightarrow\qquad y=\pm x$ Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{4}{\sqrt{8}}=\sqrt{2}$ Directrices: $\quad y=0\displaystyle \pm\frac{a}{e}=\pm\frac{\sqrt{8}}{\sqrt{2}}=\pm 2$
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