A discriminant test for the parameter orbit

Proof. Use the common model of Lemma 73.1, with base coordinates $[A:B:C]$. Its branch conic is $Q=(B^2=4AC)$. On $C=1$, put $A=r^2$, $B=2r$. The discriminant formula gives $$\Delta|_Q=(f_0-f_1r+f_2r^2-f_3r^3)^2.$$ Globally this means that the pullback of the discriminant equation to $[s:t]\mapsto[s^2:2st:t^2]$ is, up to a nonzero scalar, $(f_0t^3-f_1st^2+f_2s^2t-f_3s^3)^2$.

At a node move the point to $[0:0:1]$. The double-fibre conditions are $f_0=0$, $h_2=-f_1^2/4$. If $f_1\ne0$, the intersection multiplicity with $Q$ is two. If $f_1=0$ and the cubic form above is nonzero, it is at least four. Lemma 73.8 identifies the latter case with marked ramification, and Lemma 73.11 identifies it with vanishing sequence $(1,3)$. These are exactly the nonstable smooth pencils. The already constructed special test configurations prove strict K-semistability unless the pencil is $U_{0,0}$.

If $Q\subset\Delta$, all four $f_i$ vanish, and the equation of $X$ is $q_0+z^2h=0$. Hence $\Delta=Q\cup(h=0)$. Smoothness forces the line $(h=0)$ to be nonzero and transverse to $Q$: otherwise its intersection with $Q$ would give a nonnodal discriminant singularity, contrary to Lemma 73.3. At each of its two intersection points the marked point is ramified. The pencil therefore has two common roots of vanishing sequence $(1,3)$ and is equivalent to $U_{0,0}$ by Lemma 70.2. Conversely, that pencil has two such roots. If the displayed cubic form were nonzero, it would have a double zero at both corresponding points of $Q$, which is impossible for a cubic. Thus $Q$ is a component in this case, proving the first two alternatives exactly.

In the remaining case no common root has sequence $(1,3)$, so the pencil is GIT-stable. Its double fibres all have type $(1,2)$, and Theorem 75.7 applies. ◻