= Solution
A returning orbit must have zero net change of the <energy>. To first order, integrate the perturbation along either unperturbed homoclinic loop:
$$
0=\int_{-\infty}^{\infty}(\beta-u_h^2)v_h^2\,d\tau
=\beta I_0-I_2.
$$
For the positive loop, $v_h=\pm u_h\sqrt{a_0^2-u_h^2/2}$; the outward and inward halves contribute equally. Therefore
$$
I_0=2\int_0^{\sqrt2a_0}u\sqrt{a_0^2-u^2/2}\,du=\frac43a_0^3,
$$
and
$$
I_2=2\int_0^{\sqrt2a_0}u^3\sqrt{a_0^2-u^2/2}\,du=\frac{16}{15}a_0^5.
$$
The <integrals> follow by substituting $w=a_0^2-u^2/2$; for the second one use $u^2=2(a_0^2-w)$. Hence
$$
\boxed{\beta=\frac45a_0^2=-\frac45\alpha,\qquad
\kappa_{\rm hom}=-\frac45\lambda+o(|\lambda|),\quad\lambda<0.}
$$
The equality in rescaled parameters is the leading Melnikov balance, not an assertion that the entire finite-parameter <bifurcation> curve is exactly straight. Reflection <symmetry> makes both <saddle equilibrium> loops occur together. The global curve lies below the nonzero-equilibrium Hopf line $\kappa=-\lambda$, as shown in the parameter sketch.
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