Parabola Part 2

Concept:The image of a curve in a line is obtained by reflecting each point of the curve across that line.We use the reflection formula for a point in a line to find the image of a parametric point on the parabola, then eliminate the parameter.Explanation:Let a point on the parabola $x^2=4y$ be $(2t, t^2)$ (parametric form).The line is $x-y-1=0$.Reflection formula: for a point $(h,k)$ in line $ax+by+c=0$, the image $(x,y)$ satisfies$\frac{x-h}{a}=\frac{y-k}{b}=\frac{-2(ah+bk+c)}{a^2+b^2}$.Here $a=1$, $b=-1$, $c=-1$, $h=2t$, $k=t^2$.Apply the formula:$\frac{x-2t}{1}=\frac{y-t^2}{-1}=\frac{-2(2t - t^2 -1)}{1^2+(-1)^2} = \frac{-2(2t-t^2-1)}{2} = -(2t-t^2-1)$.Equate the first ratio: $x-2t = -(2t-t^2-1) \implies x = 2t -2t + t^2 +1 = t^2+1$.Equate the second ratio: $\frac{y-t^2}{-1} = -(2t-t^2-1) \implies y-t^2 = 2t - t^2 -1 \implies y = 2t -1$.Now eliminate $t$: from $y=2t-1$, we get $t=\frac{y+1}{2}$.Substitute into $x=t^2+1$: $x = \left(\frac{y+1}{2}\right)^2 + 1$.Simplify: $\frac{(y+1)^2}{4} = x-1 \implies (y+1)^2 = 4(x-1)$.Given image form: $(y+a)^2 = b(x-c)$.Comparing: $a=1$, $b=4$, $c=1$.Thus $a+b+c = 1+4+1 = 6$.Answer:$6$ (Option C)