Which one's are parabolas and solve?

Question

Which one's are parabolas and solve for the parabolas
1.) 4x^2+4y^2+36y+5=0
2.) 6x= y^2-6y+39
3.)y=x^2-6x+33

PUT IT IN THIS FORM y=a(x-h)^2+k or x=a(y-k)^2+h
(Whichever one is the one your supposed to use)

Answer 1

The first one is not a parabola since bth x and y are squared.

2.) 6x= y^2-6y+39
6x = (y^2-6y +9) +30
6x= (y-3)^2 +30
x = (1/6)9y-3)^2 +5

3.)y=x^2-6x+33
y = x^2-6x+9 +24
y = (x-3)^2 +24