Posted by **Dolly** on Wednesday, March 30, 2011 at 7:37am.

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

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**Reiny**, Wednesday, March 30, 2011 at 9:19am
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??

then

-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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