Question 854409: determine the vertex, directrix, and the focus of (x+4)^2 + 4(y-1), (y-2)^2 = -2(x+3), (x-3)^2 = 8(y-4), (y+1)^2 + -4(x-1), and (x-2)^2 + 4(y-2) . and graph
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! determine the vertex, directrix, and the focus of
(x+4)^2 = 4(y-1)
vertex:: (-4,1)
4p = 4, so p = 1
Basic form: y = x^2(opens upward)
directrix: y = 1-p= 0
focus: (-4,1+p) = (-4,2)
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(y-2)^2 = -2(x+3)
vertex (-3,2)
Basic form: x = -y^2(opens to the left)
4p = -2, so p = -1/2
directrix: y = -3-p = -2.5
focus: (-3+p,2) = (-3.5,2)
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Use the same procedure on the following:::
(x-3)^2 = 8(y-4), (y+1)^2 + -4(x-1), and (x-2)^2 + 4(y-2) .
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Cheers,
Stan H.
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