12y+x^2-6x+57=0; 12y+x^2-6x+9+48=0;12y+48=-(x-3)^2; y+4=-(1/12)(x-3)^2.
The equation of the parabola in this form shows that the vertex V(h,k) is at (3,-4) (y+4=0, y=-4; x-3=0, x=3). The axis of symmetry is x=3. The directrix is the line y=d (=2k-f) and the focus is at F(3,f), where d and f are to be calculated.
To find the focus, we take the coefficient of (x-3)^2 which is -1/12, which is 1/4p where p is the distance of the curve from both the focus and directrix (see section in square brackets below). So 4p=-12, making p=-3. At the vertex, the curve is equidistant from the focus and the directrix. Therefore, f=p-4=-3-4=-7 and the focus is at (3,-7). The directrix is the line y=d=-4-(-3)=-1. The parabola is an upturned, rather fat U.
To illustrate the use of focus and directrix, put x=9 in the equation of the parabola, y+4=-(1/12)36=-3, so y=-7. The point (9,-7) is 9-3=6 away from the focus and -1-(-7)=6 away from the directrix. This is as fundamental property of all parabolas: all points are equidistant from the focus and directrix line. The vertex is 3 away from the focus and directrix line.
[Derivation of focus and directrix from first principles and geometry: all points on the parabola are equidistant from the focus and directrix line:
Equation of parabola: y-k=a(x-h)^2 (y+4=-(1/12)(x-3)^2, a=-1/12, h=3, k=-4);
focus: F(h,f); directrix: y=k-(f-k)=2k-f; P(x,y); V(h,k); FV=f-k=VM, where VM is perpendicular to the directrix line.
PF=PN, where PN is perpendicular to the directrix line.
By Pythagoras: PF=(y-f)^2+(x-h)^2=(y-(2k-f))^2=PN
(y-f)^2+(y-k)/a=(y-2k+f)^2=(y+f-2k)^2
y^2-2fy+f^2+(y-k)/a=y^2+2fy+f^2-4k(y+f)+4k^2
-4fy+(y-k)/a+4k(y+f)-4k^2=0
(y-k)/a+4(-fy+ky+kf-k^2)=0=(y-k)/a+4(k(y-k)-f(y-k))=(y-k)(1/a+4(k-f))
1/a+4(k-f)=0; 1/4a+k-f=0, because y in general is not equal to k (only at the vertex)
f-k=1/4a=p (see paragraph 3 above)
p=1/4a=-3; f+4=-3, f=-7, so focus is F(3,-7).
Equation of directrix: y=-8+7=-1.]