Let the image of the parabola x^{2} = 4y in t | vedclass

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(C) The parametric point $P$ on the parabola $x^2 = 4y$ is $(2t, t^2)$.Let the mirror image of $P$ in the line $x - y - 1 = 0$ be $Q(h, k)$.Using the

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$6$

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(C) The parametric point $P$ on the parabola $x^2 = 4y$ is $(2t, t^2)$.Let the mirror image of $P$ in the line $x - y - 1 = 0$ be $Q(h, k)$.Using the reflection formula $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$, we get:$\frac{h - 2t}{1} = \frac{k - t^2}{-1} = -2 \frac{2t - t^2 - 1}{1^2 + (-1)^2} = -(2t - t^2 - 1) = t^2 - 2t + 1$.Thus, $h = 2t + t^2 - 2t + 1 = t^2 + 1$ and $k = t^2 - (t^2 - 2t + 1) = 2t - 1$.From $k = 2t - 1$, we have $t = \frac{k + 1}{2}$.Substituting $t$ into $h = t^2 + 1$, we get $h = (\frac{k + 1}{2})^2 + 1$.$h - 1 = \frac{(k + 1)^2}{4} \implies (k + 1)^2 = 4(h - 1)$.Replacing $(h, k)$ with $(x, y)$, the equation of the image parabola is $(y + 1)^2 = 4(x - 1)$.Comparing this with $(y + a)^2 = b(x - c)$, we get $a = 1, b = 4, c = 1$.Therefore, $a + b + c = 1 + 4 + 1 = 6$.

Let the image of the parabola $x^{2} = 4y$ in the line $x - y = 1$ be $(y + a)^{2} = b(x - c)$, where $a, b, c \in \mathbb{N}$. Then $a + b + c$ is equal to

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