Precalculus (6th Edition) Blitzer

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

Chapter 9 - Mid-Chapter Check Point - Page 1000: 28



Work Step by Step

Step 1. From the given conditions, we have $c=\frac{2+4}{2}=3, a=\frac{1+3}{2}=2$, thus $b^2=3^2-2^2=5$ Step 2. The center is at $(\frac{2-4}{2},5)$ or $(-1,5)$; thus we can write the equation as $\frac{(x+1)^2}{4}-\frac{(y-5)^2}{5}=1$
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