To find the image of the parabola $x^2 = 4y$ reflected across the line $L: x - y = 1$, we first determine the transformation equations for reflection.
Let $(x, y)$ be a point on the original parabola and $(x', y')$ be its image after reflection across the line $x - y = 1$.
We need to express $(x, y)$ in terms of $(x', y')$. Rearranging equations (1) and (2):
Adding these two rearranged equations:
$ (y' + x') + (x' - y') = (y + x) + (2 + y - x) $ $ 2x' = 2y + 2 \implies x' = y + 1 \implies y = x' - 1 $Subtracting the second rearranged equation from the first:
$ (y' + x') - (x' - y') = (y + x) - (2 + y - x) $ $ 2y' = 2x - 2 \implies y' = x - 1 \implies x = y' + 1 $Thus, the transformation is $x = y' + 1$ and $y = x' - 1$.
Substitute the expressions for $x$ and $y$ into the original parabola equation $x^2 = 4y$:
$ (y' + 1)^2 = 4(x' - 1) $The equation of the image parabola is $(y' + 1)^2 = 4(x' - 1)$.
We are given that the image parabola has the form $(y + a)^2 = b(x - c)$, where $a, b, c \in \mathbb{N}$.
Comparing $(y' + 1)^2 = 4(x' - 1)$ with the standard form $(y + a)^2 = b(x - c)$, we get:
Since $a = 1$, $b = 4$, and $c = 1$ are all natural numbers, this is consistent with the problem statement.
Finally, calculate the sum $a + b + c$:
$ a + b + c = 1 + 4 + 1 = 6 $