Math
posted by Dolly .
solve for the conic in the equation y^26y+12x+33=0

I am not what you mean by "solve for the conic"
your equation represents a parabola with axis parallel to the xaxis.
Are we supposed to complete the square??
then
12x = y^2  6y + 9  9 + 33
12x = (y3)^2 + 24
x = (1/12)(y3)^2  2
from that we can tell that the vertex is (2,3) and it opens to the left.