Question

**Given:** Parabola $x^2 = 4y$, line $x - y = 1$. **Parametric point** on $x^2 = 4y$: $P = (2t,\, t^2)$. **Mirror image formula:** For a point $(x_1, y_1)$ reflected in line $ax + by + c = 0$: $$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$ For line $x - y - 1 = 0$ (i.e., $a=1,\, b=-1,\, c=-1$) and point $P(2t, t^2)$: $$\frac{h - 2t}{1} = \frac{k - t^2}{-1} = \frac{-2(2t - t^2 - 1)}{2} = t^2 - 2t + 1$$ **Finding coordinates of image $Q(h, k)$:** $$h = 2t + (t^2 - 2t + 1) = t^2 + 1$$ $$k = t^2 - (t^2 - 2t + 1) = 2t - 1$$ **Eliminating parameter $t$:** From $k = 2t - 1$: $$t = \frac{k+1}{2}$$ Substituting into $h = t^2 + 1$: $$h - 1 = \left(\frac{k+1}{2}\right)^2$$ $$\therefore (k+1)^2 = 4(h-1)$$ **Locus of $Q$:** $$(y + 1)^2 = 4(x - 1)$$ **Comparing** with $(y + a)^2 = b(x - c)$: $$a = 1,\quad b = 4,\quad c = 1$$ $$\therefore a + b + c = 1 + 4 + 1 = 6$$ **Answer: (C)**