If $(3,6)$ is vertex and $(4,5)$ is focus of parabola, then the equation of directrix is

equation of $\overline{AS}$ is

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\ y-6=\frac{5-6}{4-3}(x-3)\\ y-6=\frac{-1}{1}(x-3)\\ y-6=-x+3\end{array}$

 Axis of the parabola is $x+y-9=0\to \left(1\right)$

Directrix is ${\perp }^r$ to axis of the parabola

  $\therefore x-y+k=0\to \left(2\right)$

 $A=$midpoint of $\overline{SZ}$

 $\begin{array}{l}(3,6)=(\frac{x+4}{2},\frac{y+5}{2})\\ x+4=6\\ \Rightarrow x=2\\ y+5=12\\ y=7\end{array}$

   $\therefore Z(2,7)$

 (2) passes through $Z(2,7)$

   $\begin{array}{l}2-7+k=0\\ \Rightarrow k=5\end{array}$

  $\therefore$ directrix equation is $x-y+5=0$