Question 169, Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, a line passing through the right focus $F_{2}$ intersects the ellipse at points $A$ and $B$. If $\left|F_{1} B\right|=\frac{2}{3}\left|F_{1} A\right|$, and $\angle F_{1} B A=\angle F_{1} F_{2} B$, find the eccentricity of the ellipse.
Question 169, Solution: According to the conditions, we know: $\left|\mathrm{F}_{1} \mathrm{~B}\right|=\left|\mathrm{F}_{1} \mathrm{~F}_{2}\right|=2 \mathrm{c},\left|\mathrm{F}_{1} \mathrm{~A}\right|=\frac{3}{2} \cdot\left|\mathrm{F}_{1} \mathrm{~B}\right|=\frac{3}{2} \cdot 2 \mathrm{c}=3 \mathrm{c}$. Furthermore, according to the definition of an ellipse: $\left|\mathrm{F}_{2} \mathrm{~B}\right|=2 \mathrm{a}-2 \mathrm{c},\left|\mathrm{F}_{2} \mathrm{~A}\right|=2 \mathrm{a}-3 \mathrm{c}$.
According to Stewart's Theorem:
$$
\begin{array}{l}
\left|F_{1} F_{2}\right|^{2}=\frac{\left|F_{1} A\right|^{2} \cdot\left|F_{2} B\right|+\left|F_{1} B\right|^{2} \cdot\left|F_{2} A\right|}{|A B|}-\left|F_{2} A\right| \cdot\left|F_{2} B\right| \\
\Rightarrow(2 c)^{2}=\frac{(3 c)^{2} \cdot(2 a-2 c)+(2 c)^{2} \cdot(2 a-3 c)}{(2 a-2 c)+(2 a-3 c)}-(2 a-2 c) \cdot(2 a-3 c) \\
\Rightarrow(2 e)^{2}=\frac{(3 e)^{2} \cdot(22-2 e)+\left(2 e e^{2} \cdot(2-3 e)\right.}{(2-2 e)+(2-3 e)}-(2-2 e) \cdot(2-3 e) \\
\Rightarrow(e-1)\left(5 e^{2}-11 e+4\right)=0 \\
\Rightarrow e=\frac{11-\sqrt{41}}{10}
\end{array}
$$

Therefore, the eccentricity of the required ellipse is $\frac{11-\sqrt{41}}{10}$.