= Solution
The smallest admissible axial stretch is $\lambda_{\min}=\sin2\theta$, where the two branches merge and $\alpha^2=\tan\theta=s/c$. The current unit fibre <vectors> are $m_\pm=Ul_\pm$. Their <dot product> is
$$
m_+\cdot m_-=
\frac1\lambda\left(\alpha^2c^2-\alpha^{-2}s^2\right).
$$
At the minimum, both terms in parentheses equal $cs$, so
$$
\boxed{m_+\cdot m_-=0.}
$$
Hence the two families are orthogonal at the maximum possible contraction. There is also a geometric proof of the bound: their reference parallelogram area is $\sin2\theta$, and its current area is $\sin2\theta/\lambda$, because the in-plane area stretch is $1/\lambda$. Two unit <vectors> span area at most one, with equality precisely when perpendicular. This gives $\lambda\ge\sin2\theta$ and the same equality condition without solving the quadratic. At $\theta=\pi/4$ the minimum is one, so no axial contraction is possible and the fibres are already orthogonal.
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