3. Given the parabola $C: y^{2}=2 p x(p>0)$ and points $A(a, 0), A^{\prime}(-a, 0)(t>0)$. A line passing through point $A^{\prime}$ intersects the parabola at points $P, Q$. Then the sum of the slopes of lines $A P, A Q$ is $k_{A P}+k_{A Q}=$ $\qquad$
Solution: 0.
Let points $P\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right), Q\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right)$.
From the collinearity of $A^{\prime}, P, Q$ we know
$$
\frac{y_{1}}{\frac{y_{1}^{2}}{2 p}+a}=\frac{y_{2}-y_{1}}{\frac{y_{2}^{2}}{2 p}-\frac{y_{1}^{2}}{2 p}}=\frac{2 p}{y_{1}+y_{2}} .
$$

Simplifying, we get $y_{1} y_{2}=2 p a$.
Thus, $k_{A P}=\frac{y_{1}}{\frac{y_{1}^{2}}{2 p}-a}=\frac{2 p y_{1}}{y_{1}^{2}-2 p a}=\frac{2 p y_{1}}{y_{1}^{2}-y_{1} y_{2}}=\frac{2 p}{y_{1}-y_{2}}$
Similarly, $k_{B P}=\frac{2 p}{y_{2}-y_{1}}$.
Therefore, $k_{A P}+k_{B P}=0$.