9. Let the vertex of the parabola $S$ be at the origin $O$, and its focus on the $x$-axis. A chord $AB$ is drawn through the focus $F$. The extensions of $AO$ and $BO$ intersect the directrix at $C$ and $D$, respectively. If the minimum area of quadrilateral $ABCD$ is 8, find the equation of this parabola.
9. Analysis: Let the equation of the parabola be: $y^{2}=2 p x,(p>0), A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$, and point $A$ is in the first quadrant. By the properties of the parabola, we know: $B C\|A D\| O x$, so the area of the quadrilateral $S=\frac{1}{2}\left(x_{A}+x_{B}+p\right)\left(y_{A}-y_{B}\right) ;$ Let the equation of line $A B$ be: $x=m y+\frac{p}{2}$, substituting into the parabola equation gives: $y^{2}-2 p m y+p^{2}=0$; hence $S=\frac{1}{2}\left(x_{A}+x_{B}+p\right)\left(y_{A}-y_{B}\right)=\frac{1}{2}\left(m\left(y_{A}+y_{B}\right)+2 p\right)\left(y_{A}-y_{B}\right)=2 p^{2}\left(m^{2}+1\right)^{\frac{3}{2}} \geq 2 p^{2}$, i.e., $2 p^{2}=8 \Rightarrow p=2$. Similarly, by symmetry, $p=-2$ also satisfies the condition, so the equation of the parabola is: $y^{2}= \pm 4 x$.