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Question 193559This question is from textbook algebra and trigonometry structure and method book 2
: ``Find the vertex, focus, directrix, and axis of symmetry of each parabola.
x^2+8y+4x-4=0
We have been using the formulas y-k=a(x-h)^2 and x-h=a(y-k)^2 I don't understand
which to use here, even after completing the square and its confusing because I
don't know which is which. please explain and be specific. thank you. We also
use the fact that the focus is F(h,k+c) when a=1/4c. Please explain in these
terms so that I can understand. thank you
This question is from textbook algebra and trigonometry structure and method book 2
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the vertex, focus, directrix, and axis of symmetry of each parabola.
x^2+8y+4x-4=0
We have been using the formulas y-k=a(x-h)^2 and x-h=a(y-k)^2 I don't understand which to use here, even after completing the square and its confusing because I don't know which is which. please explain and be specific. thank you. We also use the fact that the focus is F(h,k+c) when a=1/4c. Please explain in these terms so that I can understand. thank you
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x^2+8y+4x-4=0
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Complete the square on the x-terms
x^2 + 4x + ? = -8y+4 + ?
x^2 + 4x + 4 = -8y + 4+4
(x+2)^2 = -8(y-1)
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Rearrange:
(x-1) = (-1/8)(y+2)^2
h = 1, k= -2, a = -1/8 = (1/4)c
c = -1/2
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Find the vertex, focus, directrix, and axis of symmetry of each parabola.
Vertex = (h,k) = (1,-2)
Focus = (h+c, k) because this parabola is opening to the left.
Focus = (1-(1/2),-2) = ((1/2),-2)
Directrix: x = h - c = 1+1/2 = 3/2
Axis of symmetry: Horizontal line passing thru the vertex:
y = -2
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Cheers,
Stan H.
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