= Solution
The integrand has no explicit $y$ dependence. Thus the <Euler-Lagrange equation> gives the <conserved transverse optical momentum>
$$
p=\frac{\mu(x)y'}{\sqrt{1+y'^2}}.
$$
At $x=0$, $\mu(0)=\sqrt2$ and the crossing slope is one, so $p=1$. Solving for the positive slope gives $y'=1/\sqrt{\mu^2-1}$, hence
$$
\boxed{y'(x)=\begin{cases}
(1-x)^{-1/3},&-\alpha\leq x<0,\\
(1+x)^{-1/2},&0\leq x\leq\alpha.
\end{cases}}
$$
The slopes agree at zero. The same conserved momentum holds across the interface because the <refractive index> is continuous and there is no point source in the variational equation.
Here a least-time ray means the minimum between the source and the receiver point selected by this ray. For those fixed endpoints, $\partial^2[\mu\sqrt{1+y'^2}]/\partial y'^2=\mu(1+y'^2)^{-3/2}>0$ proves strict <convexity> in the slope, so the stationary ray is the unique minimum among graph paths. If the receiver could instead be freely placed anywhere on the line and its height were also minimized over, the horizontal ray would be faster; that different problem is incompatible with the prescribed crossing slope.
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