= Solution
The two halo centres orbit their <centre of mass> with separation $R$. Their relative coordinate obeys $\ddot{\mathbf R}=-G(M_1+M_2)\mathbf R/R^3$, hence circular motion requires
$$
\boxed{\omega^2=\frac{G(M_1+M_2)}{R^3}.}
$$
Reflection symmetry about the orbital plane makes the vertical force point back toward $z=0$, while the centrifugal force has no vertical component. An equilibrium away from that plane is therefore impossible.
In the uniformly rotating frame, an equilibrium is a stationary point of the gravitational plus centrifugal <effective potential>
$$
E(x,y)=-\frac{\omega^2}{2}|\mathbf r_S|^2
-\frac{GM_1}{|\mathbf r_S-\mathbf r_1|}
-\frac{GM_2}{|\mathbf r_S-\mathbf r_2|}.
$$
Set $R=1$, divide by $G(M_1+M_2)/R$, and write $\alpha=M_2/(M_1+M_2)$. The primary and secondary lie at $x=-\alpha$ and $x=1-\alpha$, so on their line
$$
F(x)=-\frac{x^2}{2}-\frac{1-\alpha}{|x+\alpha|}
-\frac\alpha{|x+\alpha-1|}.
$$
For the two roots near the secondary, put $x=1-\alpha\mp d$ in $F'(x)=0$. Dominant balance gives $3d\simeq\alpha/d^2$, so $d=(\alpha/3)^{1/3}$. Expanding the root beyond the primary directly in powers of $\alpha$ gives
$$
\boxed{L_1=(1-(\alpha/3)^{1/3},0,0),}
$$
$$
\boxed{L_2=(1+(\alpha/3)^{1/3},0,0),}
$$
$$
\boxed{L_3=(-1-5\alpha/12,0,0),}
$$
to the requested orders. The other equilibria are the two <Triangular Lagrange points>
$$
\boxed{L_{4,5}=(1/2-\alpha,\ \pm\sqrt3/2,\ 0).}
$$
The distance from $H_2$ to either nearby collinear point is the <Hill radius>. Since $\alpha\simeq M_2/M_1$,
$$
\boxed{r_t=R\left(\frac{M_2}{3M_1}\right)^{1/3}.}
$$
Inside this <tidal radius>, the subhalo's gravity dominates the host's differential gravitational field; outside it, material can escape through the neighborhoods of $L_1$ and $L_2$. For an extended spherical host, $M_1$ is replaced by enclosed mass and the coefficient becomes $3-d\log M_1/d\log R$, giving the <Jacobi tidal radius>. An extended subhalo requires the bound mass inside $r_t$ to be found self-consistently. On an eccentric orbit there is no time-independent rotating potential or exact tidal boundary; stripping is strongest near pericentre and the instantaneous radius varies around the orbit.
Because the dark component is more extended, <tidal stripping> first sends dark matter through both $L_1$ and $L_2$, producing leading and trailing dark-matter <tidal tails>. The compact stellar component is stripped more deeply and also produces a leading and a trailing stellar tail. The two constituents therefore give four tails distinguished by composition, with the dark tails broader and more extended.
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