Indeed it does - nicely done.
Putting it into a more familiar format of y(x) instead of the paired equations of y(b) and x(b).
Coordinates of the apex being, as you solved it,
x = -b/(2a)
y = -b^2/(4a) + c
Squaring the first equation gives x^2 = b^2/(4a^2).
Tweaking the second equation and then substituting x into it gives
y = -b^2/(4a) + c = -a * b^2 / (4a^2) + c = -ax^2 + c
So it seems not just any parabola, but identical in shape with the original, upside down, and has the apex at (0,c) which is where all the parabolas y = ax^2 + bx + c pass through the y-axis.