Question 426979
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Hi
 Using The standard form is {{{(x -h)^2 = 4p(y -k)}}}, where the focus is (h,k + p) 
Find standard form of parabola when directix is y=4 and focus is (6,-6)
Note the vertex is 'halfway' (along the line of symmetry, x = 6) 
between the directrix: y = 4 and the focus (6,-6).(this parabola opens downward)
distance between directrix and focus: 4-(-6) = 10 , 10/2 = 5...  
p = -5... therefore, the Vertex Pt is (6,-1)
 (x-6)^2 = -20(y+1)  OR y = (-1/20)(x-6)^2 - 1
{{{drawing(300,300, -10, 10, -10, 10,blue(line(6,10,6,-10)),  grid(1),
circle(6, -1,0.4),
circle(6, -6,0.4),
graph( 300, 300,  -10, 10, -10, 10,0,4,(-1/20)(x-6)^2 - 1))}}}