SOLUTION: Please help me find the axis of symmetry, the focus, directrix and vertex of the parabola. (x+2) ^ 2 - 16 (y-1) = 0 thank you!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me find the axis of symmetry, the focus, directrix and vertex of the parabola. (x+2) ^ 2 - 16 (y-1) = 0 thank you!      Log On


   

Question 733804: Please help me find the axis of symmetry, the focus, directrix and vertex of the parabola.
(x+2) ^ 2 - 16 (y-1) = 0
thank you!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find the axis of symmetry, the focus, directrix and vertex of the parabola.
(x+2) ^ 2 - 16 (y-1) = 0
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Form of parabola opening up:
(x+2)^2 = 16(y-1)
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Vertex: (-2,1)
Axis of symmetry: x = -2
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Since 4p = 16, p = 4
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Directrix: y = -3
Focus:: (-2,5)
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16y-16 = (x+2)^2
16y = (x+2)^2+16
y = (1/16)(x+2)^2+1
graph%28400%2C400%2C-10%2C10%2C-10%2C10%2C%281%2F16%29%28x%2B2%29%5E2%2B1%29
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Cheers,
Stan H.
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