Solution

Parametric point $ P $ on $ x^2 = 4y $ is $ P(2t, t^2) $

$ \therefore $ mirror image of $ P $ in $ x - y = 1 $ is

$ Q = \left( \frac{2t - 2\cdot1\cdot(2t - t^2 - 1)}{2}, \frac{2t^2 - 2\cdot1\cdot(2t - t^2 - 1)}{2} \right) $

$ Q = (t^2 + 1,; 2t - 1) = (h, k) $

$ \Rightarrow $ locus of $ Q $ is $ x = \frac{(y + 1)^2}{4} - 1 $ which is required parabola

$ \therefore (y + 1)^2 = 4(x - 1) $

$ \therefore a = 1,; b = 4,; c = 1 $

$ \therefore a + b + c = 6 $