Rearrange: x^2+6x+9-9-51=12y; (x+3)^2=12(y+5).
Take a point P(x,y), focus at F(h,k) and directrix line is y=d. Properties of a parabola are such that the distance from any point on the curve is equal to the perpendicular distance of the point to the directrix. So we can write:
(Distance between P and F)^2=(x-h)^2+(y-k)^2.
Perpendicular distance PQ from P to the directrix=y-d. PQ=PF, so PQ^2=PF^2
PQ^2=(y-d)^2=(x-h)^2+(y-k)^2, from which (y-d)^2-(y-k)^2=(x-h)^2 and (k-d)(2y-d-k)=(x-h)^2. Compare this with the rearranged equation and we can see that (x-h)^2=(x+3)^2 so h=-3.
Now the y term. 2y(k-d)=12, so k-d=6.
The constant term is 60=(d-k)(k+d)=-6(k+d), so k+d=-10. Therefore, 2k=-4 and k=-2 and d=-8.
The focus is (-3,-2) and the directrix is y=-8.
The focus lies on the vertex line so x=-3 is the vertex line. When x=-3, y=-5, so the vertex is at (-3,-5).
Check: (x-h)^2+(y-k)^2=(y-d)^2; (x+3)^2+(y+2)^2=(y+8)^2 should be true for all (x,y) on the curve. So, (x+3)^2=12(y+5), which is the rearranged equation above, and if we insert the vertex coords we can see it's correct.