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I have a situation for which I have made a very crude drawing. Let's say we have an ellipse in $\mathbb{R}^2$ that is fixed at $x_0 = -a$ and $x_1 = a$ (as if it were resting on two poles). I am curious to know how we can discuss the relative motion of the curve if we were to move one of the "poles" away such that $x_1^{'} = x_1 + \Delta x$ for some $\Delta x \in \mathbb{R}$.

Any assistance would be greatly appreciated.

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  • $\begingroup$ Is the point on the ellipse touching the left post fixed and the ellipse merely rotate around that point? Or instead is there some form of gravity where the ellipse can slide to touch the two posts? $\endgroup$ – David G. Stork Apr 9 at 18:59
  • $\begingroup$ There is some form of gravity (assuming the normal form). The points are initially fixed. Then one point moves over and gravity would act. I assume we can consider things like the principle of least action but I am not sure on the formalism. $\endgroup$ – JacobCheverie Apr 9 at 19:05
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This is an outline of a solution to your problem. I’m going to rename some of your quantities to conform to conventions about equations of an ellipse. We have an ellipse $(x/a)^2+(y/b)^2=1$, the distance $d$ of the pivot point $P$ from the $y$-axis and the moving point $Q$. The coordinates of these points are $$P = \left(-d,-\frac ba\sqrt{a^2-d^2}\right) \\ Q=\left(d+\Delta,\frac ba\sqrt{a^2-d^2}\right) = (-x_P+\Delta, y_P).$$ As $Q$ moves horizontally, the ellipse pivots around $P$ to maintain contact with $Q$. The point of contact on the unrotated ellipse satisfies $PR=PQ$, i.e., it is the other end of a chord from $P$ with length $PQ$. This suggests that $R$ can be found by computing the intersection of a circle with the ellipse. We’ll come back to this later. Since the segment $\overline{PQ}$ is horizontal, it’s fairly easy to work out that the rotation angle $\theta$ satisfies $$\cos\theta = {x_R-x_P\over x_Q-x_P}, \sin\theta = -{y_R-y_P\over x_Q-x_P}, \tan\theta = -{y_R-y_P\over x_R-x_P}.$$ The center of the rotated ellipse can be found in various ways to be $$x_c = x_P-x_P\cos\theta+y_P\sin\theta \\ y_c = y_P-y_P\cos\theta-x_P\sin\theta$$ and an equation of the rotated ellipse is therefore $${(x\cos\theta-y\sin\theta-x_c)^2\over a^2}+{(x\sin\theta+y\cos\theta-y_c)^2\over b^2}=1.$$

All that’s left is to compute point $R$ and then do some tedious algebra to express the above in terms of $d$ and $\Delta$. The general solution to the system $${x^2\over a^2}+{y^2\over b^2}=1 \\ (x+d)^2+\left(y+\frac ba\sqrt{a^2-d^2}\right)^2=(2d+\Delta)^2$$ is quite messy, though, and there’s still the matter of selecting the correct solution. It might be worth exploring parameterizing by the distance between $P$ and $Q$ instead of the the offset $\Delta$ from the original position of $Q$ to see if it simplifies any of the resulting expressions.

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