Lemma 10.28.3.DFE3E0889355 The strict transforms of the conics
\[ F_1-2F_2=z(x-2y)+txy=0 \]
form a surface $\Gamma \subset \mathcal X$. For $t\ne 0$, its fiber $C_ t$ is a smooth compact rational curve with $L_ t\cdot C_ t=0$. The central fiber of $\Gamma $ is $\ell +q$, where $q$ is the strict transform of the line $\{ x=2y\} $. The curves $\ell $ and $q$ meet transversely at one point $p$, and $L_0\cdot q=1$. Near every point of $q\setminus \{ p\} $, $\Gamma $ is smooth and $\pi |_\Gamma $ is a submersion.
Proof.
The symmetric matrix of the quadratic form has determinant $-t/2$, so the conic is smooth for $t\ne 0$. It passes simply through all four centers, so its strict transform is a smooth rational curve of class $L_ t$, and its degree against $L_ t$ is $L_ t^2=0$.
At $t=0$ the conic is the union of the lines $\{ z=0\} $ and $\{ x=2y\} $. The second line avoids $p_1,p_2,p_3$ and passes through $p_4$. Each component has multiplicity one at every center it contains, so the central fiber of $\Gamma $ contains no additional exceptional component. The two lines meet at $[2:1:0]$, transversely and away from every center. Since $q=H-E_4$, we have $L_0\cdot q=1$.
Away from the centers, the defining equation at $t=0$ has nonzero derivative in the directions of $\mathbb P^2$ at every point of $q\setminus \{ p\} $, which proves the assertion there. Near $p_4$, use the blowup chart $z=1$, $x=r$, $y=rs$. After the exceptional factor is removed, the equation becomes
\[ 1-2s+trs=0. \]
At the central point of $q\cap E_4$ we have $r=0$ and $s=1/2$, and the derivative in $s$ is $-2$. Thus $(r,t)$ are local coordinates on $\Gamma $ there. These two cases cover $q\setminus \{ p\} $.
$\square$
Pome
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