Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 709: 30

Answer

Hyperbola Foci: $( 1,\pm \sqrt 2)$ Vertices: $(1,\pm1,)$

Work Step by Step

$x^2-2x=y^2-2$ $\frac{(y-0)^{2}}{1^2}-\frac{(x-1)^{2}}{1^2}=1$, the equation of a hyperbola. $c^{2}=1+1=2$ $c=\sqrt 2$ Foci: $( 1,\pm \sqrt 2)$ Vertices: $(1,\pm1)$
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.