Let the equation of the tangent P on the parabola ${\mathrm{x}}^2-4\mathrm{x}-4\mathrm{y}+16=0$ be $2\mathrm{x}-\mathrm{y}-5=0$. If the equation of the normal drawn at P to this parabola is $\mathrm{ax}+\mathrm{y}+\mathrm{c}=0,$then ac =

Equation of the tangent to parabola ${\mathrm{x}}^2-4\mathrm{x}-4\mathrm{y}+16=0$ is

$x{x}_1-2\left(x+x_1\right)-2\left(y+y_1\right)+16=0$

$\left(x_1-2\right)x-2y+\left(-2x_1-2y_1+16\right)=0···\left(i\right)$

and given tangent is $2\mathrm{x}-\mathrm{y}-5=0···\left(ii\right)$

Equation (i) and (ii) represents same line then

$\frac{x_1-2}{2}=\frac{-2}{-1}=\frac{\left(-2x_1-2y_1+16\right)}{-5}$

$x_1=6,y_1=-3$