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Question 54271: find the vertex, focus, directrix of the parabola by x^2 -4x -6y -16 = 0
Found 2 solutions by Nate, stanbon:
Answer by Nate(3500)   (Show Source):
You can put this solution on YOUR website! x^2 - 4x - 16 = 6y
x^2 - 4x = 6y + 16
(x - 2)^2 = 6y + 20
(x - 2)^2 - 20 = 6y
(1/6)(x - 2)^2 - 10/3 = y
vertex: (2,-10/3)
p = 1/(4a) = 1/(4/6) = 3/2 = 1.5
foci: (2,-10/3 + 1.5)
directrix: y = -10/3 - 1.5
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! find the vertex, focus, directrix of the parabola by x^2 -4x -6y -16 = 0
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Rewrite as:
x^2-4x+? = 6y+16+?
Complete the square:
x^2-4x+4=6y+16+4
(x-2)^2 = 6(y+(10/3))
This follows the form (x-h)^2=4p(y-k)
h=2; k=-10/3; p=3/2
Vertex at (2,10/3); focus=(2,(10/3)+(3/2))=(2,29/6);
directrix: y = (10/3)-(3/2)=11/6
Cheers,
Stan H.
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