SOLUTION: write the equaton of a parabola that has a focus at (-1,7), has a minimum and the length from focus to vertex is 2 units

Since it has a minimum, it is of the form

(x-h)2 = 4a(y-k)

and the vertex is below the focus, and therefore opens upward.

Since the length from focus to vertex is 2 units,

1. the vertex is 2 units below the focus (-1,7), so

the vertex is (h,k) = (-1,5)

and 

2.  |a| = the distance from focus to vertex = 2, positive since
    the parabola opens upward.

So the equation

(x-h)2 = 4a(y-k)

becomes

(x-(-1))2 = 4(2)(y-(5))
 
(x+1)2 = 8(y-5)

Edwin