Question 70893
Determine an equation of the parabola with the following property 
1) Focus at (0,-6) and directrix y=6.
Plot the point and sketch the line on a coordinate system.
The vertex is half way between the directrix and the focus.
So the vertex is (0,0)
"p" is the distance from the vertex to the focus:
p=-6
Therefore 4p=-24
So the equation of the parabola is:
(x-0)^2 = -24(y-0)
x^2 = -24y
y=(-1/24)x^2
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THEN it says 
determine the vertex,focus,and directrix for each parabola. Sketch each parabola. 
1) y[squared]=-3x 
Put the equation in vertex form which is (y-k)^2 = 4p(x-h)
(y-0)^2 = -3(x-0)
vertex is (0,0)
4p = -3
p = -3/4
So the focus is at (-3/4,0)
The directrix is x=3/4
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2)x[squared]=12y 
(x-0)^2 = 12(y-0)
vertex = (0,0)
4p=12
p=3
Focus is at (0,3)
Directrix is y=-3
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3) x[squared]-8x-y+20=0 
x^2-8x = y-20
Complete the square to get:
x^2-8x+16 = y-20+16
(x-4)^2 = y-4
vertex at (4,4)
4p = 1
p= 1/4
Focus (4,4+1/4) or (4,17/4)
directrix at y=4-(1/4) 
y=15/4
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Cheers,
Stan H.