Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.7 - Conics in Polar Coordinates - Exercises 11.7 - Page 686: 20

Answer

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

Work Step by Step

$y^{2}-x^{2}=4\qquad /\div 4$ $\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{4}=1\qquad $ We have a vertical axis with, $a=2, b=2$ Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{4+4}=2\sqrt{2}$ Foci: $\quad (0, \pm c)=\qquad (0, \pm 2\sqrt{2})$ Vertices: $\quad (0, \pm a)=\qquad (0, \pm 2)$ Asymptotes: $\quad y=\displaystyle \pm\frac{a}{b}x\Rightarrow\qquad y=\pm x$ Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{2\sqrt{2}}{2}=\sqrt{2}$ Directrices: $\quad y= y=0\displaystyle \pm\frac{a}{e}=\pm\frac{2}{\sqrt{2}}=\pm\sqrt{2}$
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.