The Chebyshev polynomial of the first kind is defined on $[-1, 1]$ by
$$T_n(x) = \cos(n \arccos x).$$
Prove that the envelope for the extremals of $T_{n+1}(x)-T_{n-1}(x)$ forms an ellipse.
The conclusion is obvious from the picture, and we know that the extremals are "almost" but not lying on $x^2+\frac{y^2}{4}=1$, but I think it is not easy to strictly illustrate this property and derive the final conclusion (The extremal is the zeros of $x\sin(n\cos^{-1}(x))+n\sqrt{1-x^2}\cos(n\cos^{-1}(x))=0$, and I think the "envelope" is somewhat hard to describe). Are there any ways to strictly prove the correctness of this problem? Thanks!