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: 41

Answer

$\frac{(x+1)^{2}}{12}+\frac{(y-4)^{2}}{16}=1$,

Work Step by Step

Distance between the center $(-1,4)$ and focus $(-1,6)$ is 2, which is the value of $c$. $a^2=b^2-c^2=16-4=12$ $\frac{(x+1)^{2}}{12}+\frac{(y-4)^{2}}{16}=1$, This is an equation of an ellipse.
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