Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 16

Answer

Foci: $( -1\pm \sqrt 2,2 )$ Vertices : $( -1 \pm \sqrt 3,2)$

Work Step by Step

$x^2+3y^2+2x-12y+10=0$ Rewrite as $\frac{(x+1)^2}{(\sqrt 3)^2}+\frac{(y-2)^2}{1}=1$ $c^2=a^2-b^2=3-1=2$ $c=\sqrt 2$ Foci: $( -1\pm \sqrt 2,2 )$ Vertices : $( -1 \pm \sqrt 3,2)$
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