Question 743917
Given (x-4)^2 = 4(y+1)  ------------------> (1)

Comparing to (x-h)^2 = 4a(y-k) ------------------> (2)

(h,k) = (4,-1) ----------> Vertex 

From (1) & (2),

4a = 4 ,[ a=1 ]

Equation of directrix y-k=-a,  y+1=-1, y=-2 --------> Equation of directrix

Vertex of Parabola (x-h=0, y-k=a)
x=h     ,    y=a+k
x=4     ,    y=0
(4,0) ---------------------------------> Vertex of Parabola.

Cheers