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solve for the conic in the equation y^2-6y+12x+33=0

  • Math - ,

    I am not what you mean by "solve for the conic"

    your equation represents a parabola with axis parallel to the x-axis.
    Are we supposed to complete the square??

    -12x = y^2 - 6y + 9 - 9 + 33
    -12x = (y-3)^2 + 24
    x = (-1/12)(y-3)^2 - 2

    from that we can tell that the vertex is (-2,3) and it opens to the left.

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