Lemma 10.28.8.49F835066112 For $a\in \mathbb C\setminus \{ 0,1\} $, the strict transforms of the conics
\[ F_1-aF_2=z(x-ay)+t(a-1)xy=0 \]
have smooth rational fibers of $L_ t$-degree zero for $t\ne 0$. Their central fiber is $\ell +q_ a$, where $q_ a=f^{-1}(a)$ is a smooth projective line of $L_0$-degree one. The two components meet transversely at $p_ a=[a:1:0]$, away from the blowup centers. Along $q_ a\setminus \{ p_ a\} $, the surface formed by these conics is smooth and its projection to the disk is a submersion.
Proof.
The symmetric matrix of the quadratic form has determinant $-a(a-1)t/4$, so the conic is smooth for $t\ne 0$; it passes simply through the four centers. Its strict transform has class $L_ t$, so its degree is $L_ t^2=0$. At $t=0$ the conic splits into the two lines $\{ z=0\} $ and $\{ x=ay\} $. Since $a$ is finite and $a\notin \{ 0,1\} $, the line $\{ x=ay\} $ does not pass through the other three central points, so no additional exceptional component occurs, and $q_ a=H-E_4$ has degree one against $L_0$. The two lines meet transversely at $p_ a$, away from all centers.
Away from the centers, the central equation has nonzero derivative in the directions of $\mathbb P^2$ along $q_ a\setminus \{ p_ a\} $. At the point of $q_ a$ on $E_4$, the chart $z=1$, $x=r$, $y=rs$ gives, after the exceptional factor is removed,
\[ 1-as+t(a-1)rs=0. \]
At $r=t=0$ and $s=1/a$, the derivative in $s$ is $-a\ne 0$. These charts prove the final assertion.
$\square$
Pome
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