= Solution
Only the upper <equilibrium> can have zero <trace> with positive <determinant>. Setting $s_+=\lambda$ gives
$$
\boxed{\mu_H(\lambda)=\lambda(\lambda-1),\qquad \lambda>\frac12.}
$$
At this point the <determinant> is $\lambda^2(2\lambda-1)>0$, so the angular frequency is $\lambda\sqrt{2\lambda-1}$. Moreover $ds_+/d\mu=1/\sqrt{1+4\mu}>0$, so the <trace> crosses zero transversely as $\mu$ increases. The stipulated <supercritical Hopf bifurcation> creates a stable <periodic orbit> on the side $\mu>\mu_H$ where the upper <equilibrium> becomes unstable.
The steady-state curves are the vertical lines $\mu=-1/4$ and $\mu=0$ in the $(\mu,\lambda)$ <plane>. The Hopf parabola has its endpoint at
$$
\boxed{(\mu,\lambda)=(-1/4,1/2),\qquad (u,v)=(1/2,1/4).}
$$
There the Jacobian has <trace> and <determinant> zero but is not the zero matrix: it has a nilpotent Jordan block. This is the codimension-two Takens-Bogdanov point. The intersection of the Hopf curve with $\mu=0$ at $\lambda=1$ concerns a different <equilibrium> from the transcritical crossing at the origin; it is not another double-zero <equilibrium>.
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