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

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

The correct answer is

Finding the Reflection Transformation

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 $x - y = 1$.

The midpoint $M = \left(\frac{x+x'}{2}, \frac{y+y'}{2}\right)$ must lie on the line $x - y = 1$. $ \frac{x+x'}{2} - \frac{y+y'}{2} = 1 \implies x + x' - y - y' = 2 \quad \quad (1) $
The line segment connecting $(x, y)$ and $(x', y')$ must be perpendicular to the line $x - y = 1$. The slope of $x - y = 1$ is $1$. The slope of the segment is $\frac{y' - y}{x' - x}$. $ \frac{y' - y}{x' - x} = -1 \implies y' - y = -(x' - x) \implies y' - y = -x' + x \quad \quad (2) $

Solving for Original Coordinates

We need to express $(x, y)$ in terms of $(x', y')$. Rearranging equations (1) and (2):

From (2): $y' + x' = y + x$
From (1): $x' - y' = 2 + y - x$

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 $y = x' - 1$.

Substituting into the Parabola Equation

Substitute the expressions for $x$ and $y$ into the original parabola equation $x^2 = 4y$:

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

Identifying Parameters $a, b, c$

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:

$y + a = y' + 1 \implies a = 1$
$b = 4$
$x - c = x' - 1 \implies c = 1$

Since $a = 1$, $b = 4$, and $c = 1$ are all natural numbers, this is consistent with the problem statement.

Calculating $a + b + c$

Finally, calculate the sum $a + b + c$:

$ a + b + c = 1 + 4 + 1 = 6 $

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