Question

Let the image of 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

• A. 4
• B. 6
• C. 12
• D. 8

Hint:

To find the image of a curve under reflection, transform the coordinates first and then substitute into the original equation.

Answer 1

The Correct Option is B

To find the image of the parabola \( x^2 = 4y \) under reflection in the line \( x-y=1 \), we will follow several steps. The line can be rewritten as \( y = x - 1 \). The process involves the following steps:

1. Finding the Image of the Vertex:

The vertex of the parabola \( x^2 = 4y \) is at the origin \((0,0)\). To find its image, we use the reflection formula for a point \((x_1, y_1)\) in a line \(Ax + By + C = 0\):

1. \(x' = \frac{x_1(B^2 - A^2) - 2A(By_1 + C)}{A^2 + B^2}\) \(y' = \frac{y_1(A^2 - B^2) - 2B(Ax_1 + C)}{A^2 + B^2}\)

Substituting \((x_1, y_1) = (0, 0)\) into the formula for the line \(x-y-1=0\) (where \(A=1\), \(B=-1\) and \(C=-1\)), simplifies the reflection. The image of \((0,0)\) is:

1. \((0,0) \to (1, -1)\)
2. Transformation of the Parabola:

The general equation of a parabola reflected over a line is found by transforming each point \((x, y)\) to its image. This step requires applying reflection to a general point on the parabola and finding the corresponding new parabola equation.

Post-reflection, the equation \(y^2 = 4x\) (derived from the standard form using the image vertex) maintains the standard parabola form. Therefore, we can write the image of the original parabola as \((y+1)^2 = 4(x-1)\).

1. Identifying Constants:

The transformed parabola can be expressed as \((y+a)^2 = b(x-c)\), where the vertex \((1, -1)\) gives \(a=1\), \(b=4\), and \(c=1\).

Thus, \(a + b + c = 1 + 4 + 1 = 6\).

1. Conclusion:

Therefore, the sum \(a+b+c\) is equal to 6.