11. As shown in Figure 1, the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ has left and right foci $F_{1}$ and $F_{2}$. Points $A$ and $B$ are on the upper half of the ellipse, and $F_{1} A \parallel F_{2} B$. Let $B F_{1}$ and $A F_{2}$ intersect at point $P$. Find the maximum value of the y-coordinate of point $P$.
11. As shown in Figure 5, let $\theta=\angle x F_{2} B, r_{1}=\left|A F_{1}\right|, r_{2}=\left|C F_{1}\right|$, where $C$ is the other intersection point of $A F_{1}$ with the ellipse.
According to the focal radius formula, we have
$$
\begin{array}{l}
r_{1}=\frac{3}{2-\cos \theta}, r_{2}=\frac{3}{2+\cos \theta} \\
\Rightarrow \frac{1}{r_{1}}+\frac{1}{r_{2}}=\frac{4}{3} .
\end{array}
$$

Draw $P Q / / A F_{1} / / B F_{2}$ intersecting the $x$-axis at point $Q$.
$$
\begin{array}{l}
\text { By } \frac{|P Q|}{\left|A F_{1}\right|}=\frac{\left|Q F_{2}\right|}{\left|F_{1} F_{2}\right|}, \frac{|P Q|}{\left|B F_{2}\right|}=\frac{\left|F_{1} Q\right|}{\left|F_{1} F_{2}\right|} \\
\Rightarrow \frac{|P Q|}{\left|A F_{1}\right|}+\frac{|P Q|}{\left|B F_{2}\right|}=1 . \\
\text { Also, }\left|B F_{2}\right|=\left|C F_{1}\right|=r_{2} \text {, then } \\
\frac{|P Q|}{r_{1}}+\frac{|P Q|}{r_{2}}=1 \\
\Rightarrow|P Q|=\frac{1}{\frac{1}{r_{1}}+\frac{1}{r_{2}}}=\frac{3}{4} .
\end{array}
$$

Thus, $y_{p} \leqslant|P Q|=\frac{3}{4}$, and the equality holds when $\theta=\frac{\pi}{2}$.
Therefore, the maximum value of the $y$-coordinate of point $P$ is $\frac{3}{4}$.