If we have to find normal to the $x^2 = 4by$
I tried
Let a point be $(-2bm,bm^2)$
Then equation of tangents is
$-2bmx = 2b(y + bm^2)$ hence slope is $-2m$
So the slope of normal is $1/(2m)$ and equation is
$Y-bm^2 = (\frac{1}{2}m) ( x + 2bm)$
But this is not correct
I will assume that you were asked to find the normal at the point $(-2bm,bm^2)$.
From $x^2=4by$ we find that $\frac{dy}{dx}=\frac{2x}{4b}$. When $x=-2bm$ we have $\frac{dy}{dx}=-m$.
So the slope of the normal, if $m\ne 0$, is $\frac{1}{m}$.
The normal therefore has equation $y-bm^2=\frac{1}{m}\left(x+2bm\right)$. You may be expected to give a "simplified" form, such as $y=\frac{x}{m}+2b+bm^2$.