By “sliding normally” I mean that the vertex of the parabola is a point $P$ that circulates around the ellipse, and that the axis of the parabola is normal to the tangent (to the ellipse) at $P$.
https://www.desmos.com/calculator/vewhhesehw
Let a point $P= (f(a), g(a))=( 6 \cos(a), 3 \sin(a))$ (with $a$ a varying angle) move on the ellipse $ x^2/6 + y^2 / 3^2 = 1$.
My goal is to find the cartesian equation of a parabola (1) similar to $y=x^2$ but (2) the vertex of which is located at $P$ and (3) the axis of which is normal to the tangent (to the ellipse) passing through $P$, and finally (4) that opens outwards.
Saying that the axis is normal to the tangent passing through $P$ means that the inclination of the parabola is the angle $ R = \arctan \frac {g'(a)} {f'(a)} = \arctan \left(-\frac {3\cos(a)} {6\sin(a)} \right)$.
Using the “rotation about any point” formula, I first arrived at
$$X(x,y)= (x- 6 \cos(a)) \cos(R) + ( y - 3\sin(a))\sin(R)$$
$$Y(x,y)= ( y- 3\sin(a))\cos(R) - ( x - 6 \cos(a)) \sin(R)$$
which yields, for the parabola,
$$Y(x,y) = (X(x,y))^2 \space\space \Bigg( \space \iff Y(x,y) - X(x,y)^2 =0 \Bigg)$$
The problem is that, with the above formula, the parabola opens inwards when $P$ is below the $X$ axis.
The only way I found to fix this problem was $\DeclareMathOperator{\sgn}{sgn}$
(1) to put $\sgn(a)$ next to $Y(x,y)$
(2) and to limit the range of $a$ from $-\pi$ to $+\pi$, which gives:
$$Y(x,y)\sgn(a) = { (X(x,y))^2} \space\space \Bigg( \space \iff Y(x,y)\sgn(a) - X(x,y)^2 =0 \Bigg)$$
Can you find a way to solve the above problem without using these palliatives?