## Precalculus (6th Edition) Blitzer

$(x-4)^2=12(y-2)$
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 $(x-4)^2=12(y-2)$