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: 39

Answer

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Work Step by Step

The polar equation for a conic with eccentricity $e$ is $\displaystyle \quad r=\frac{ke}{1+e\cos\theta},$ where $x=k \gt 0$ is the vertical directrix, and $\left\{\begin{array}{ll} e=1 & \Rightarrow\text{parabola}\\ e \lt 1 & \Rightarrow\text{ellipse}\\ e \gt 1 & \Rightarrow\text{hyperbola} \end{array}\right.$ $\displaystyle \frac{25\div 10}{(10-5\cos\theta)\div 10}=\frac{\frac{5}{2}}{1-(\frac{1}{2})\cos\theta}=\frac{-5(-\frac{1}{2})}{1+(-\frac{1}{2})\cos\theta}$ So, $e=\displaystyle \frac{1}{2},\qquad \Rightarrow\text{ellipse}$ and $\quad k=5 \Rightarrow x=-5$ is a directrix The origin is the left focus. $ r=\displaystyle \frac{a(1-e^{2})}{1+e\cos\theta}\quad$ is the polar equation of an ellipse. $a(1-e^{2})=\displaystyle \frac{5}{2}$ $a(1-\displaystyle \frac{1}{4})=\frac{5}{2}$ $a\displaystyle \cdot\frac{3}{4}=\frac{5}{2}$ $a=\displaystyle \frac{10}{3}$ $c=ea=\displaystyle \frac{1}{2}\cdot\frac{10}{3}=\frac{5}{3}$ The center of the ellipse is $\displaystyle \frac{5}{3}$ units right of the focus at the origin. In polar coordinates, at $(\displaystyle \frac{5}{3},0).$ The vertices are at $\displaystyle \frac{10}{3}$ units left/right of the center; in polar coordinates, $(\displaystyle \frac{5}{3}+\frac{10}{3},0)=(5,0)$ and $(\displaystyle \frac{5}{3}-\frac{10}{3},0)=(-\frac{5}{3},0)$ or $(\displaystyle \frac{5}{3},\pi)$
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