9. As shown in the figure, given the parabola $C: y^{2}=4 x$, a chord $A P, A Q$ of the parabola $C$ is drawn through the point $A(1,2)$.
(1) If $A P \perp A Q$, prove: the line $Q P$ passes through a fixed point, and find the coordinates of the fixed point;
(2) Assuming the line $P Q$ passes through the point $T(5,-2)$, does there exist an isosceles triangle $A P Q$ with $P Q$ as the base?

If it exists, find the number of $\triangle Q A P$, if not, explain the reason.
9. Analysis: Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and the line $P Q: x=m y+n$.
(1) Since $P A \perp Q A$, we have: $\left(x_{1}-1\right)\left(x_{2}-1\right)+\left(y_{1}-2\right)\left(y_{2}-2\right)=0$, simplifying to: $\left(m^{2}+1\right) y_{1} y_{2}+(m n-m-2)\left(y_{1}+y_{2}\right)+n^{2}-2 n+5=0(*)$. By combining the parabola and the line $P Q$ equations, we get: $y^{2}-4 m y-4 n=0$. Using Vieta's formulas, $y_{1}+y_{2}=4 m, y_{1} y_{2}=-4 n$. Substituting into $(*)$ gives $(n-3)^{2}-4(m+1)^{2}=0$, hence $n=2 m+5$ or $n=-2 m+1$. Therefore, the line $P Q: x=m y+2 m+5$ or $x=m y-2 m+1$, passing through the fixed points $(5,-2)$ and $(1,2)$, respectively. Since the line $P Q$ clearly does not pass through point $A$, it passes through the fixed point $(5,-2)$.
(2) When the line $P Q$ passes through the fixed point $(5,-2)$, from the conclusion of part (1), we know $P A \perp Q A$. At this time, the equation of the line $P Q$ is $x=m y+2 m+5$. When $\triangle Q A P$ is an isosceles triangle with $P Q$ as the base, $|A P|=|A Q|$, so $\left(x_{1}-1\right)^{2}+\left(y_{1}-2\right)^{2}=\left(x_{2}-1\right)^{2}+\left(y_{2}-2\right)^{2}$. Simplifying gives $m\left(x_{1}+x_{2}-2\right)+\left(y_{1}+y_{2}\right)-4=0(* *)$. From the conclusion of part (1), we know $y_{1}+y_{2}=4 m$. Substituting into $(* *)$ gives $m^{3}+m^{2}+3 m-1=0$. Considering the function $f(x)=x^{3}+x^{2}+3 x-1$, $f^{\prime}(x)=3 x^{2}+2 x+3 \geq 0$ always holds, so $m^{3}+m^{2}+3 m-1=0$ has only one real root. Therefore, there is one such isosceles triangle.