Let the image of parabola \(x^2=4 y\), in the line \(x-y=1\) be \((y+a)^2=b(x-c), a, b, c \in N\). Then \(a+b+c\) is equ…

Parametric point \(P\) on \(x^2=4 y\) is \(P\left(2 t, t^2\right)\)

∴ mirror image of \(P\) in \(x-y=1\) is

\(Q\equiv \left(2t−\frac{2⋅1⋅\left(2t−{t}^{2}−1\right)}{2},{t}^{2}+\frac{2⋅1⋅\left(2t−{t}^{2}−1\right)}{2}\right)\)

\(Q\equiv \left({t}^{2}+1,2t−1\right)\equiv (h,k)\)

So, \(h=t^2+1, k=2t-1\)

\(\Rightarrow t=\frac{k+1}{2}\)

\(\Rightarrow h= \frac{(k+1)^2}{4} +1\)

∴ locus of \(Q\) is \(x=\frac{{(y+1)}^{2}}{4}+1\) which is the required parabola.

\(∴{(y+1)}^{2}=4\left(x−1\right)\)

\(∴a=1,b=4,c=1\)

\(∴a+b+c=6\)