Question 172, Given the hyperbola \( C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0), F_{1}, F_{2} \) are the left and right foci of \( C \), respectively, and \( P \) is a point on the right branch of \( C \) such that \( \angle F_{1} \mathrm{PF}_{2}=\frac{\pi}{3} \), and the area of \( \triangle \mathrm{F}_{1} \mathrm{PF}_{2} \) is \( 3 \sqrt{3} \mathrm{a}^{2} \).
(1) Find the eccentricity \( e \) of \( C \);
(2) Let \( A \) be the left vertex of \( C \), and \( Q \) be any point on \( C \) in the first quadrant. Does there exist a constant \( \lambda (\lambda>0) \) such that \( \angle \mathrm{QF}_{2} \mathrm{~A}=\lambda \angle \mathrm{QAF}_{2} \) always holds? If it exists, find the value of \( \lambda \); if not, explain the reason.
Question 172, Solution: (1) Let $F_{1}(-c, 0), F_{2}(c, 0), P(x, y)$, then according to the focal radius formula, we have $\left\{\begin{array}{l}\left|\mathrm{PF}_{1}\right|=\mathrm{ex}+\mathrm{a} \\ \left|\mathrm{PF}_{2}\right|=\mathrm{ex}-\mathrm{a}\end{array}\right.$ . According to the conditions:
$$
\begin{array}{l}
\left\{\begin{array}{c}
(e x+a)^{2}+(e x-a)^{2}-(e x+a)(e x-a)=(2 c)^{2} \\
\frac{(e x+a)(e x-a) \sin \frac{\pi}{3}}{2}=3 \sqrt{3} a^{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{c}
e^{2} x^{2}+3 a^{2}=4 c^{2} \\
e^{2} x^{2}=13 a^{2}
\end{array}\right. \\
\Rightarrow e=\frac{c}{a}=2
\end{array}
$$
(2) According to the analysis in (1), the hyperbola C: $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{3 \mathrm{a}^{2}}=1$. And $A(-a, 0), F_{2}(2 a, 0)$. If there exists a $\lambda$ that satisfies the condition, such that for any point $Q$ in the first quadrant of $\mathrm{C}$, we have $\angle \mathrm{QF}_{2} \mathrm{~A}=\lambda \angle Q A F_{2}$. First, take $Q(2 a, 3 a)$, then $\angle Q F_{2} A=90^{\circ}, \angle Q A F_{2}=45^{\circ}$, so $\lambda=2$. Therefore, $\angle \mathrm{QF}_{2} \mathrm{~A}=2 \angle Q A F_{2}$. Furthermore, generally for any $Q$, we have $\tan \angle \mathrm{QF}_{2} \mathrm{~A}=\frac{\mathrm{y}_{0}}{2 \mathrm{a}-\mathrm{x}_{\mathrm{Q}}}, \tan 2 \angle Q A F_{2}=\frac{2 \tan \angle Q A F_{2}}{1-\tan ^{2} \angle Q A F_{2}}=\frac{2 \frac{y_{Q}}{x_{Q}+a}}{1-\left(\frac{y_{Q}}{x_{Q}+a}\right)^{2}}$. Hence
$$
\angle \mathrm{QF}_{2} \mathrm{~A}=2 \angle \mathrm{QAF}_{2} \Leftrightarrow \frac{\mathrm{y}_{Q}}{2 \mathrm{a}-\mathrm{x}_{Q}}=\frac{2 \frac{y_{Q}}{x_{Q}+3}}{1-\left(\frac{y_{Q}}{x_{Q}+3}\right)^{2}} \Leftrightarrow \frac{x_{Q}^{2}}{\mathrm{a}^{2}}-\frac{y_{Q}^{2}}{3 \mathrm{a}^{2}}=1
$$

This shows that $\lambda=2$ satisfies the condition.