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