Making a coordinate system slide around a parabola in order to ( hopefully) make a parabola slide on its fixed coordinate system. The sate of my research using Desmos :https://www.desmos.com/calculator/t5w5ocxauf
My final goal is to find the equation of a parabola that is slidind on the X-axis of its coordinat system.
My idea is to reverse the question, i.e. to make a coordinate system slide on the parabola, and then to rewrite the equation of the parabola with respect to the formerly sliding system, taken as fixed.
The parabola I use if defined by $ \large f(x)= \frac 1 5 x^2$
I take the moving $Y$ axis to be the tangent to the parabola at a moving point $P= ( a, f(a))$. The equation of this tangent is :
$y = f'(a) ( x-a) + f(a) = \frac {2a} {5} (x-a)+ \frac {2 a^2} {5}$
The moving X axis should satisfy 2 conditons :
(1) it has to be normal to the tangent , that is, its slope has to be :
$ - \frac {1} { f'(a)} =  - \frac {1} { 2a/5} = - \frac {5} {2a}$
(2) it has to touch the parabola at a point $ Q= (b, f(b)$, that is ,its slope also
has to be  : $f'(b)= \frac {2b} {5}$.
It follows that $ - \frac {5} {2a} = \frac {2b} {5} \implies  b = \frac {-25} {\space 4a} \implies Q= \big(\frac {-25} {\space 4a}, f( \frac {-25} {\space 4a}\big) \big)$
Consequently, the equation of the moving $X$ axis is ( in terms of number $a$)  :
$\large y = -\frac {1} {f'(a)} ( x+ \frac {25} {\space 4a}) + f( \frac {-25} {\space 4a})$
https://www.desmos.com/calculator/t5w5ocxauf

In which direction should I go now?
Should I work on the orientation of the two lines, that is their
inclination using the arctan function? An try to deduce the
inclination of the moving parabola from the inclination of the alledgedly
sliding coordinate system?
Should I try to find a ( parametric ? ) equation for the movement of the two contact
points on the movng axis, and try to translate,  after that , these
equatons for the fixed system ? In hat case, the sliding  parabola (
in the fixed system) will have to be tangent to the axes at these
moving contact points?

 A: Your goal is to slide a parabola on the axes in the first quadrant.  The parabola is defined in its own coordinate system as
$ y' = f(x') = \dfrac{1}{5} x'^2 $
And $O'x'y'$ is related to the fixed world coordinate system by
$ (x, y) = (x_0, y_0) + R (x', y') $
where $R$ is the rotation matrix
$ R = \begin{bmatrix} \cos(\theta) && - \sin(\theta) \\ \sin(\theta) && \cos(\theta) \end{bmatrix} $
Now you just need to find the location of the origin $(x_0, y_0)$ as a function of $\theta$, so that the parabola touches the $x$-axis and $y$-axis
The parametric equation of the parabola in world coordinates is
$ (x(t), y(t) ) = (x_0 + \cos(\theta) t - \frac{1}{5} \sin(\theta) t^2 , y_0 + \sin(\theta) t + \frac{1}{5} \cos(\theta) t^2 ) $
The tangent vector is the derivative of this parametric expression with respect to $t$.
$ T(t) = ( \cos(\theta) - \frac{2}{5} \sin(\theta) t , \sin(\theta) + \frac{2}{5} \cos(\theta) t ) $
If $T(t)$ is pararllel to the $x$ axis, then $T_y(t) = 0 $, from which
$ t = - \frac{5}{2} \tan(\theta) $
At this value of $t$ we want $y(t)$ to be zero as well.  Hence, we require that
$ y_0 -\frac{5}{4} \sin(\theta)\tan(\theta) = 0 $
And this gives us $ y_0 = \frac{5}{4} \sin(\theta) \tan(\theta) $
Similarly, if our parabola is to touch the $y$ axis then $T_x(t) = 0 $ for some $t$, at which
$ \cos(\theta) - \frac{2}{5} \sin(\theta) t = 0 $
From which , $ t = \frac{5}{2} \cot(\theta) $.
Now use this $t$ in the expression for $x(t)$, you get,
$ x_0 + \cos(\theta) t - \frac{1}{5} \sin(\theta) t^2 = 0 $
So that
$ x_0 = -\frac{5}{4} \cos(\theta) \cot(\theta) $
The result is on this desmos page
