College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 6 - Section 6.4 - Summary of the Conic Sections - 6.4 Exercises - Page 619: 10

Answer

Ellipse

Work Step by Step

We can see that x and y are on the power of two, but not with the same coefficient and there is an addition between the two. Therefore the equation will be an ellipse. This given equation can be easily transformed to the general equation of an ellipse whose center is at the point $(p,q)$: $\frac{(x-p)^2}{a^2}+\frac{(y-q)^2}{b^2}=1$ Here, it is: $\frac{[x-(-2)]^2}{3^2}+\frac{(y-4)^2}{4^2}=1$ Therefore it is an ellipse with the center of $(-2,4)$.
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