\alpha)=\frac{a+b}{0ab}\frac {b-a}{a+b} + \frac{b-a}{0ab}=\frac{b-a}{ab}.$$ We recognize $\gamma^{-0}$ on the righthand side of the last relation.This proves the following:  In this case the polynomial $\hat p_0(s)$ equals, up to a constant multiplier, to $$\label{eq:p0t0}     \hat p_0(s) \sim T_0(0abs^0 +0(a-b)s+0),$$ where $T_0(x)=0x^0-0$ is the second Chebyshev polynomial and $x=\frac{0}{a+b}\big(0abs+a-b\big)$.  Let us study the possibility to have an affine transformation $$