Let, $\left(h,k\right)$ be the image of the point $\left({\lambda }^2,2\lambda \right)$ about the line $x-y+1=0$

$\frac{h-{\lambda }^2}{1}=\frac{k-2\lambda }{-1}=\frac{-2\left({\lambda }^2-2\lambda +1\right)}{2}$
$h+1=2\lambda \dots ..\left(\mathrm{i}\right)$
$k={\lambda }^2+1\dots ..\left(\mathrm{i}\mathrm{i}\right)$
Put the value of $\lambda$ from $\left(\mathrm{i}\right)$ in $\left(\mathrm{i}\mathrm{i}\right)$
$k-1={\left(\frac{h+1}{2}\right)}^2$

$\Rightarrow 4\left(y-1\right)={\left(x+1\right)}^2$
Hence, $a=-1, b=4$ and $c=1$