Question

Let mirror image of parabola $x^2 = 4y$ in the line $x-y=1$ be $(y+a)^2 = b(x-c)$. Then the value of $(a+b+c)$ is

• A. 3
• B. 6
• C. 9
• D. 12

Hint:

For mirror image problems, parametric points simplify reflection calculations greatly.

Answer 1

The Correct Option is B

Step 1: Take a parametric point on the parabola.
For $x^2 = 4y$, a parametric point is \[ P = (2t,\,t^2) \] Step 2: Find mirror image of point in the line $x-y=1$.
Using reflection formula, mirror image $Q(h,k)$ of $P$ is \[ Q = \left(2t - \frac{2(2t - t^2 - 1)}{2},\; t^2 + \frac{2(2t - t^2 - 1)}{2}\right) \] Step 3: Simplify coordinates.
\[ Q = (t^2 + 1,\,2t - 1) \] Step 4: Eliminate parameter $t$.
From $y = 2t - 1$, \[ t = \frac{y+1}{2} \] Substitute in $x = t^2 + 1$, \[ x = \frac{(y+1)^2}{4} + 1 \] Step 5: Write equation in standard form.
\[ (y+1)^2 = 4(x-1) \] Step 6: Compare with given equation.
\[ (y+a)^2 = b(x-c) \] So, \[ a = 1,\quad b = 4,\quad c = 1 \] Step 7: Final calculation.
\[ a+b+c = 1+4+1 = 6 \] Final conclusion.
The value of $(a+b+c)$ is 6.