By " sliding nomally" I mean that the axis of the parabola is normal to the tangent to the $ f(x)= 3 cos(x/3)$ curve at point $(a, f(a))$.
The question seems to amount to doing 3 things
(1) translating the coordinate system to the new center $(a, f(a))$, which yields :
$$y-f(a) = (x-a) ^2$$
(2) rotating the coordinate system centered at $(a, f(a))$ by an angle $\alpha = arctan ( f'(a))$, and rewriting the parabola in this rotated system :
$$(y- f(a)) cos(\alpha) - (x-a) sin(\alpha) = [ (x-a) cos (\alpha) + (y-f(a)) sin (\alpha) ] ^2 $$
(3) rewriting $y= x^2$ in the original coordinate system :
$$(y- f(a)) cos(\alpha) - (x-a) sin(\alpha) + \underline {f(a)} = [ (x-a) cos (\alpha) + (y-f(a)) sin (\alpha) + \underline{a} ] ^2 $$
But this does not work. The orientation of the transformed parabola is apparently OK, but the location of the vertex is not as desired. Where does my reasoning go wrong?
I aimed at following the procedure discribed in this post How to rotate curve around a point.
Desmos link https://www.desmos.com/calculator/ijk9l2shqr
A parabola is rotated away from vertical by introducing xy and y^{2} terms into the general equation. There is only one parabola equation. It is $4p(y-k)=(x-h)^{2}.$ However, we can substitute a set of axis rotations for $x$ and $y$ and in so doing, we create two new terms in the general equation. The substitutions are $$x=x'\cos\theta+y'\sin\theta$$
$$y=-x'\sin\theta+y'\cos\theta.$$ We might as well account for the $(h,k)$ vertex offset at the same time. That makes the substitution be
$$\left(\begin{array}{c}
x=(x'-h)\cos\theta+(y'-k)\sin\theta\\
y=-(x'-h)\sin\theta+(y'-k)\cos\theta.
\end{array}\right)$$
Now the parabola equation becomes $$4p\left[-(x'-h)\sin\theta+(y'-k)\cos\theta\right]=(x'\cos\theta+y'\sin\theta)^{2}.$$
It is easiest to let the substitutions be their own functions.
$$\left(\begin{aligned}a(x,y)= & (x-h)\cos\theta+(y-k)\sin\theta\\
b(x,y)= & -(x-h)\sin\theta+(y-k)\cos\theta
\end{aligned}
\right)$$
Then the parabola equation is $$4p\left(a(x,y)\right)=\left(b(x,y)\right)^{2}$$
In your specific case: $f(x)=3\cos(x/3)$$V=(q,f(q))$
slope: $m=-1/f^\prime (q)$ where $f^\prime(x)=-\sin\left(\frac{1}{3}x\right)$
$\theta = \arctan(m)$
$a(x,y)=(x - x(V)) \cos(θ) + (y - y(V)) \sin(θ)$
$b(x,y)=-(x - x(V)) \sin(θ) + (y - y(V)) \cos(θ)$
$a(x, y) = (b(x, y))²$ is the rotated and translated parabola
[![parabola image][1]][1]