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Question 170128: im familar with writing conic sections in there standard equation but there are 2 questions that have me stumped in peticular.here is one of them.
y^2-8y-8x+64=0
and im suppose to write this in standard form
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! y^2-8y-8x+64=0
and I'm supposed to write this in standard form
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Complete the square of the y terms, as follows:
y^2 - 8y + 16 = 8x -64 + 16
(y-4)^2 = 8x - 48
(y-4)^2 = 8(x-6)
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This form shows you the vertex is at (6,4) and the
focus of the parabola is 8/4 = 2 units above the vertex.
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Cheers,
Stan H.
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