Question

What is the equation of the quadratic graph with a focus of (5, −1) and a directrix of y = 1?

Answer 1

(x-h)^2=4p(y-k)
vertex is (h,k)
p is the distance from focus to vertex and distance from vertex to directix (vertex is in middle of directix and focus)
if p is positive, the parabola opens up and the focus is above the directix
if p is negative, the parabola opens down and the focus is below the directix
 
we see the directix is over the focus (1>-1) so the parabola opens down and p is negative
distance from (5,-1) to y=1 is 2 units
2/2=1
p=-1 since p is negative
1 up from (5,-1) is (5,0)
veretx is (5,0)
(x-5)^2=4(-1)(y-0)
(x-5)^2=-4y is the equation

Answer 2

The answer is directrix is horizontal, so the parabola is vertical 
focus lies below the directrix, so the parabola opens downwards. 

General equation for down-opening parabola: 
 y = a(x-h)²+k 
with 
 a<0 
 vertex (h,k) 
 focal length p = 1/|4a| 
 focus (h,k-p) 
 directrix (h,k+p) 

Apply your data and solve for h, k, and a. 
vertex is halfway between focus and directrix: (3,3) 
h = 3 
k = 3 

distance between focus and vertex is 2, so p = 2. 
a = -1/(4p) = -⅛ 

y = -⅛(x-3)²+3