Quite a straightforward piece of maths I can't seem to get my head around here:
The tangent to the parabola at the point $(4a, 4a)$ is given by what equation?
Bearing in mind the parabola is $y^2 = 4ax$ and the general point $(at^2, 2at)$ lies on the parabola.
Do you know how to compute the tangents to $y=\frac{1}{4a}x^2$ by differentiation? If we take the point $(t,\frac{1}{4a}t^2)$, the tangent has equation $$ y-\frac{1}{4a}t^2=\frac{1}{2a}t(x-t) $$ because the derivative of $x\mapsto \frac{1}{4a}x^2$ is $x\mapsto \frac{1}{2a}x$. For $t=4a$, we get $$ y-\frac{1}{4a}16a^2=\frac{1}{2a}4a(x-4a) $$ that is $$ y-4a=2x-8a $$ or $$ y=2x-4a $$
Just swap $x$ with $y$ and the requested tangent will be $$ x=2y-4a $$
Note that swapping $x$ with $y$ is an isometry, so it preserves tangents.
Let's make another example. Suppose you want to compute the normal to the parabola at the point $(a,-2a)$. You can do it in a very similar way, by computing the normal to the parabola $y=\frac{1}{4a}x^2$ at the point $(-2a,a)$. The tangent will be $$ y-a=\frac{1}{2a}\cdot(-2a)(x+2a) $$ or $$ y-a=-x-2a $$ that is $$ y=-x-a $$ The normal is thus $$ y=x-a $$ so the required normal is $$ x=y-a $$ (by swapping back $x$ with $y$).