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Question 98399: I need the following please:
standard form for: x^2-4x-2y-8=0
vertex and focus of the following parabola: y^2+10y-x+3=0
find the equation of the directrix of the parabola: 4x^2-8x-2y-4=0
thanks for the help
I have worked them - just not sure
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! standard form for: x^2-4x-2y-8=0
2y = x^2-4x-8
y = (1/2)x^2-2x-4
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vertex and focus of the following parabola: y^2+10y-x+3=0
x = y^2+10y+3
Complete the square on the y terms:
x-3+? = y^2+10y+?
y^2+10y+25 = x-3+25
(y+5)^2 = x+22
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Vertex: (-22,-5)
4p=1 so p=1/4
The parabola opens to the right so
focus = (-22+(1/4),-5) = (-87/4,-5)
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find the equation of the directrix of the parabola: 4x^2-8x-2y-4=0
Divde thru by 2 to get:
2x^2-4x-y-2=0
y+2 = 2x^2-4x
y+2+2*? = 2(x^2-2x+?)
2(x^2-2x+1) = y+2+2
2(x-1)^2 = y+4
(x-1)^2 = (1/2)(y+4)
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Vertex = (1,-4)
4p=(1/2)
p = 1/8
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Parabola opens up so directrix is p below the vertex.
Directrix: y = -4-(1/8) = -33/8
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Cheers,
Stan H.
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