The equation is of type #(y-k)^2=a(x-h)#, which has #y-k=0# as axis of symmetry and vertex is #(h,k)#. Focus is #(h+a/4,k)# and directrix is #x=h-a/4#
The given equation #y^2=-12x# is of this type as we can write it as #(y-0)^2=-12(x-0)#. Compare #y^2=-12x# with #(y-k)^2=a(x-h)#, we have #h=0#, #k=0# and #a=-12#.
Hence, axis of symmetry is #y=0# and vertex is #(0,0)#. Further, focus and vertex lie on axis of symmetry and directrix is perpendicular to it. Hence equation of directrix is of type #x=b#,
Focus is #(0+(-12)/4,0)# or #(-3,0)# and
directrix is #x=-(-12)/4=3#
graph{(y^2+12x)((x+3)^2+y^2-0.03)(x-3)=0 [-10, 10, -5, 5]}
