Solution
= Solution
Writing the horizontal coordinate as $u=\log\lambda$, so that $\lambda=e^u$, the two coefficient paths at $\alpha=1/2$ are
$$
\beta_1(u)=\frac{(1.41-e^u/2)_+}{1+e^u/2},
\qquad
\beta_2(u)=-\frac{(0.41-e^u/2)_+}{1+e^u/2}.
$$
Equivalently, as functions of $\lambda$, replace every $e^u$ by $\lambda$. The paths reach zero at $lambda=2.82$ and $lambda=0.82$, respectively.