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

Step 1: Reflection formula across $x-y=1$.
Reflection of point $(x,y)$ across line $x-y=1$ gives \[ (x',y')=(1-y,1-x) \] Step 2: Substitute in the original parabola.
Original equation: \[ x^2=4y \] Replace $x=1-y'$ and $y=1-x'$: \[ (1-y')^2=4(1-x') \] Step 3: Simplify.
\[ y'^2-2y'+1=4-4x' \] \[ (y'-1)^2=4(x'-1) \] Step 4: Compare with standard form.
\[ (y+a)^2=b(x-c) \] Here, \[ a=-1,\quad b=4,\quad c=1 \] Step 5: Compute required sum.
\[ a+b+c=1+4+1=6 \]