This question is related to a question I asked a while ago here on math.stackexchange: Moving Line Segment Problem
The rules for how the line segment can be moved are the same: The endpoints must stay on the curves they started on, and the length of the line segment must be preserved through the entirety of the movements of the endpoints along the given curves.
This time, however, the curves are different. Instead of using a pair of circles, we will have the functions $f(x)=-x^2+1$ and $g(x)-2x^2+2$.
The starting position of the line segment will have the endpoint on the curve defined by $f$ lying in $[-1,0)$ and the endpoint on the curve defined by $g$ will be in $(-1,0]$ and the endpoint on $g$ will be closer to $0$ than the endpoint on $f$.
There are two questions here:
1) For what lengths, $l$, can this line segment exist, and be moved to the other side?
2) For what lengths among the answers for 1) can it be moved to the other side in such a way that it ends up precisely where it's reflection over the $y$-axis would be?
Edit: It was brought to my attention that I failed to communicate one thing about the movement of the endpoints: I intended for them to also be constrained in $x$ by the requirement that $x\in[-1,1]$ for both endpoints at all times.
I'll have a stab at this. First an image of the functions:
1) The length of the line segment must be $l\ge 1$
It cannot be less than 1 because the minimum distance at $x = 0$ is 1. And it doesn't need to be greater than 1 as the minimum distance between the curves decreases in the rest of the interval $[-1,1]$.
Regarding maximum length, suppose the line segment has an endpoint on $g(x)$ in the $[-1,0[$ interval (left of the top) and the endpoint of f(x) is also left of the top. When the endpoint of g(x) passes the top ($x = 0$) the endpoint on f(x) can just move backward ($x_f$ decreasing) and continue moving backward until the line segment is horizontal. At which point the endpoint on f(x) starts moving forward again. Hence no maximum.
BTW, I'm assuming that the line segment must only remain attached to both curves while at least 1 endpoint has $-1\le x\le1$. Otherwise there is no solution as the parabolas diverge indefinitely and no length would be enough.
2) The only length which would give a symmetrical move is $l = 1$
If $l>1$ then when the endpoint on g(x) is at the top of g(x) ($x = 0$), the endpoint on f(x) must be either on the left or right side of the top of f(x). And as the minimum distance decreases to either side, the endpoint on f(x) can only move backward until the endpoint on g(x) has moved far enough to allow forward movement. This means the movement across the top will not be symmetrical.
I realize the above is not a proof. Let me know if and where the reasoning fails.