Answer
$(x-4)^2=12(y-2)$
Work Step by Step
Step 1. From the given conditions, we have $2p=5+1=6$ or $p=3$
Step 2. The center is at $(4,\frac{5-1}{2})$ or $(4,2)$; thus we can write the equation as $(x-4)^2=4(3)(y-2)$ or $(x-4)^2=12(y-2)$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 9 - Mid-Chapter Check Point - Page 1000: 29
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