The equation of the normal to the parabola, x^2=8 y at x = 4 is

Putting x = 4 in $x^2=8 y$, we get y = 2. Now,

$\Rightarrow x^2=8 y \Rightarrow 2 x=8 \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{x}{4}$

$\Rightarrow \left(\frac{d y}{d x}\right)_{(4,2)}=1$

The equation of the normal at (4, 2) is

$y-2=-\frac{1}{1}(x-4)$ or, $x+y=6$