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: 21

Answer

$\displaystyle \frac{x^{2}}{2}-\frac{y^{2}}{8}=1$ Foci: $\qquad (\pm\sqrt{10},0)$ Directrices: $\quad x=\displaystyle \pm\frac{\sqrt{10}}{5}$

Work Step by Step

Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$ Write in standard form and read a,b. $8x^{2}-2y^{2}=16\qquad /\div 16$ $\displaystyle \frac{x^{2}}{2}-\frac{y^{2}}{8}=1\qquad $ We have a horizontal axis with $a=\sqrt{2}, b=2\sqrt{2}$ Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{2+8}=\sqrt{10}$ Foci: $\quad (\pm c, 0)=\qquad (\pm\sqrt{10},0)$ Vertices: $\quad (\pm a, 0)=\qquad (\pm\sqrt{2}, 0)$ Asymptotes: $\quad y=\displaystyle \pm\frac{b}{a}x \Rightarrow \quad y=\pm 2x$ Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{5}$ Directrices: $\quad x=0\displaystyle \pm\frac{a}{e}=\pm\frac{\sqrt{2}}{\sqrt{5}}=\pm\frac{\sqrt{10}}{5}$
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