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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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