$\because \quad$ Vertex $\mathrm{O} \equiv(2,3)$ focus $\mathrm{S} \equiv(3,2)$ $\because \mathrm{O}$ is the mid point AS $\therefore \frac{\mathrm{x}_1+3}{2}=2$ and $\frac{\mathrm{y}_1+2}{2}=3$
$\Rightarrow x_1=1, y_1=4$
Slope of AS, $m_1=\frac{2-3}{3-2}=-1$
$\because$ Directrix is perpendicular to axis of parabola:
$\Rightarrow \mathrm{m}_2=1 \quad\left\{\because \mathrm{m}_2\right.$ is slope of directrix $\}$
$\left(y-y_1\right)=m_2\left(x-x_1\right) \Rightarrow y-4=1(x-1)$
$\Rightarrow \quad y-x=3$ ...(i)
Let point $\mathrm{P}$ is on parabola
Since, $M P=P S$
$\left\{\because\right.$ as $\mathrm{M}$ is foot of $\perp^{\mathrm{r}}$ from $\mathrm{P}$ to the directrix. $\}$
$\begin{aligned} & \Rightarrow \frac{y-x-3}{\sqrt{1+1}}=\sqrt{(x-3)^2+(y-2)^2} \\ & \Rightarrow(y-x-3)^2=2\left((x-3)^2+(y-2)\right)^2 \\ & \Rightarrow x^2+y^2-18 x-2 y+2 x y+17=0\end{aligned}$