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: 24

Answer

$\displaystyle \frac{x^{2}}{36}-\frac{y^{2}}{64}=1$ Foci: $\qquad (\pm 10,0)$ Directrices: $\quad x=\displaystyle \pm\frac{18}{5}$

Work Step by Step

Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$ Write in standard form and read a,b. $64x^{2}-36y^{2}=2304\qquad /\div 2304$ $\displaystyle \frac{x^{2}}{36}-\frac{y^{2}}{64}=1\qquad $ We have a horizontal axis, $a=6, b=8$ Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{36+64}=10$ Foci: $\quad (\pm c, 0)=\qquad (\pm 10,0)$ Vertices: $\quad (\pm a, 0)=\qquad (\pm\sqrt{2}, 0)$ Asymptotes: $\quad y=\displaystyle \pm\frac{b}{a}x \Rightarrow \quad y=\displaystyle \pm\frac{4}{3}x$ Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$ Directrices: $\quad x=0\displaystyle \pm\frac{a}{e}=\pm\frac{6}{\frac{5}{3}}=\pm\frac{18}{5}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.