Question 171, Given the ellipse $C: \frac{x^{2}}{3}+y^{2}=1$ with its upper vertex at $A$, a line $l$ not passing through $A$ intersects the ellipse $C$ at points $P$ and $Q$, and $\overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{AQ}}=0$. (1) Prove that the line $l$ always passes through a fixed point; (2) Try to find the maximum possible area of $\triangle \mathrm{APQ}$.
Question 171, Solution: (1) According to the conditions, $A(0,1)$. Let the line $1: y=kx+b$, then $b \neq 1 \%$ and the system of equations is $\left\{\begin{array}{l}y=kx+b \\ \frac{x^{2}}{3}+y^{2}=1\end{array}\right.$, then
$$
\begin{array}{l}
\frac{x^{2}}{3}+(kx+b)^{2}=1 \\
\Rightarrow\left(3k^{2}+1\right)x^{2}+6kbx+3\left(b^{2}-1\right)=0
\end{array}
$$

According to Vieta's formulas, we have: $\left\{\begin{array}{l}x_{P}+x_{Q}=-\frac{6kb}{3k^{2}+1} \\ x_{P} \cdot x_{Q}=\frac{3\left(b^{2}-1\right)}{3k^{2}+1}\end{array}\right.$.
Since $\overrightarrow{AP} \cdot \overrightarrow{AQ}=0$, hence
$$
\begin{array}{l}
0=x_{P} \cdot x_{Q}+\left(y_{P}-1\right)\left(y_{Q}-1\right)=x_{P} \cdot x_{Q}+\left(kx_{P}+b-1\right)\left(kx_{Q}+b-1\right) \\
=x_{P} \cdot x_{Q}+k^{2}x_{P} \cdot x_{Q}+k(b-1)\left(x_{P}+x_{Q}\right)+(b-1)^{2} \\
=\frac{3\left(k^{2}+1\right)\left(b^{2}-1\right)}{3k^{2}+1}-\frac{6k^{2}b(b-1)}{3k^{2}+1}+(b-1)^{2} \\
=\frac{2(2b+1)(b-1)}{3k^{2}+1}
\end{array}
$$

Thus, we know $b=-\frac{1}{2}$. This indicates that the line 1 always passes through the point $\left(0,-\frac{1}{2}\right)$.
(2) Noting that $b=-\frac{1}{2}$, hence the line $1: y=kx-\frac{1}{2}$, and $\left\{\begin{array}{c}x_{P}+x_{Q}=\frac{3k}{3k^{2}+1} \\ x_{P} \cdot x_{Q}=-\frac{9}{4\left(3k^{2}+1\right)}\end{array}\right.$. Therefore, we have:
$$
\begin{array}{l}
=\frac{3\left|x_{Q}-x_{P}\right|}{4}=\frac{3 \sqrt{\left(x_{P}+x_{Q}\right)^{2}-4x_{P} \cdot x_{Q}}}{4}=\frac{3 \sqrt{\left(\frac{3k}{3k^{2}+1}\right)^{2}-4\left(\frac{9}{4\left(3k^{2}+1\right)}\right)}}{4} \\
=\frac{9 \sqrt{4k^{2}+1}}{4\left(3k^{2}+1\right)} \leq \frac{9}{4} \\
\end{array}
$$

When $k=0$, the equality holds.