Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.1 - The Ellipse - Concept and Vocabulary Check - Page 965: 8

Answer

$4$, $1$, $16$

Work Step by Step

Rewrite the equation as $3(x^2+4x+4)+4(y^2-2y+1)=32+12+4$ or $3(x+2)^2+4(y-1)^2=48$ This means that we complete the square on $x$ by adding $4$ within the first parentheses. Similarly, we complete the square on $y$ by adding $1$ within the second parentheses. At the same time, we must add $16$ to the right side of the equation.
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