Question 1015625
What's the standard form of the equation of the parabola at 
1.) Vertex at (0, 5), directrix is the line x=-2
With those conditions the parabola must be opening to the right::
p = -2
(y-k)^2 = 4p(x-h)
(y-5)^2 = -8x 
y^2 - 10y + 25 = -8x
x = (-1/8)y^2 + (5/4)y - (25/8)

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2.) Vertex at (6,6), directrix is the line y=7
With those conditions the parabola must be opening down
p = 1
(x-h)^2 = 4*1(y-k)
(x-6)^2 = 4(y-6)
x^2-12x+36 = 4y-24
4y = x^2 - 12x + 60
y = (1/4)x^2 - 3x + 15
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Cheers,
Stan H.
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