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 685: 6

Answer

$F(\pm 1,0)$ Eccentricity$:\qquad e =\displaystyle \frac{1}{\sqrt{10}}$ Directrices$:\quad x=\pm 10$ Graph:

Work Step by Step

$9x^{2}+10y^{2}=90\qquad /\div 90$ $\displaystyle \frac{x^{2}}{10}+\frac{y^{2}}{9}=1,\qquad a=\sqrt{10},b=3$ The major axis is horizontal. $c=\sqrt{a^{2}-b^{2}}=\sqrt{10-9}=1$ $F(\pm 1,0)$ Eccentricity$:\qquad e=\displaystyle \frac{c}{a}=\frac{1}{\sqrt{10}}$ Directrices$:\displaystyle \quad x=0\pm\frac{a}{e}$ $x=\displaystyle \pm\frac{\sqrt{10}}{\frac{1}{\sqrt{10}}}$ $x=\pm 10$
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.