Question 719762: Why are the vertex, directrix, and focus of the parabola x^2-6x-8y+49=0 Found 2 solutions by lynnlo, stanbon:Answer by lynnlo(4176) (Show Source):
You can put this solution on YOUR website! Why are the vertex, directrix, and focus of the parabola x^2-6x-8y+49=0
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Solve for "y":
8y = x^2 -6x + 49
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y = (1/8)x^2 - (3/4)x + (49/8)
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Complete the square:
y = (1/8)[x^2 - [(3/4)/(1/8)]x + ?] + (49/8) - (1/8)*?
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y = (1/8)[x^2 - (6x) + 9] + (49/8)-(9/8)
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y = (1/8)(x-3)^2 + 5
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Vertex: (3,5)
4p = 8
p = 2
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Focus: (3,7)
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Directrix: y = 1
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Cheers,
Stan H.
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