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

$x+y=6$ $x+2 y=0$ $3-2 y=0$ $x+y=2$

$x+y=6$

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, $x+y=6$