7. Let $P$ be a moving point on the ellipse $\frac{y^{2}}{4}+\frac{x^{2}}{3}=1$, and let points $A(1,1), B(0,-1)$. Then the maximum value of $|P A|+|P B|$ is $\qquad$ .
Answer: 5. Solution: Take $F(o, 1)$, then $F, B$ are the upper and lower foci of the ellipse, respectively. By the definition of an ellipse, $|P F|+|P B|=4$. Therefore, $|\mathrm{PA}|+|\mathrm{PB}|=4-|\mathrm{PF}|+|\mathrm{PA}| \leqslant 4+|\mathrm{FA}|=4+l=5$.
When $\mathrm{P}$ is at the intersection of the extension of $\mathrm{AF}$ and the ellipse $\left(-\frac{3}{2}, 1\right)$, $|\mathrm{PA}|+|\mathrm{PB}|$ reaches its maximum value of 5.