Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.4 - The Parabola; Identifying Conic Sections - Exercise Set - Page 799: 58

Answer

Hyperbola

Work Step by Step

Since we are given $4x^2=36+y^2$ It can be written as: $4x^2-y^2-36=0$ Thus, this shows that the variable $x^2$ and $y^2$ having opposites signs, so the graph of the given equation is a hyperbola.
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