Point of contact of the line kx + y – 4 = 0 w.r.t. the parabola y = x – x² is

$\begin{array}{l}\mathrm{kx}+\mathrm{y}-4=0\Rightarrow \mathrm{y}=-\mathrm{kx}+4\\ sub y in y=x-x^{2 }then x^2-(k+1)x+4=0\\ \Rightarrow \mathrm{\Delta }=0\Rightarrow \mathrm{k}+1=\pm 4\Rightarrow \mathrm{k}=-5 or 3\\ line becomes 3x+y-4=0 or -5x+y-4=0\\ by verfication point (2,-2) lies on both line and parabola\end{array}$