12. In the Cartesian coordinate system $x O y$, it is known that points $A_{1}(-2,0), A_{2}(2,0)$, and a moving point $P(x, y)$ satisfies the product of the slopes of lines $A_{1} P$ and $A_{2} P$ is $-\frac{3}{4}$. Let the trajectory of point $P$ be curve $C$.
(1) Find the equation of $C$;
(2) Suppose point $M$ is on the line $x=4$, and two lines passing through $M$ intersect $C$ at points $A, B$ and $P, Q$ respectively, and $|M A| \cdot|M B| = |M P| \cdot|M Q|$. Find the sum of the slopes of lines $A B$ and $P Q$.
(1) $k_{A_{1} P} \cdot k_{A_{2} P}=\frac{y^{2}}{x^{2}-4}=-\frac{3}{4} \Rightarrow \frac{y^{2}}{3}+\frac{x^{2}-4}{4}=0 \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let $M(4, s), A B:\left\{\begin{array}{l}x=4+t \cos \alpha, \\ y=s+t \sin \alpha,\end{array} P Q:\left\{\begin{array}{l}x=4+t \cos \beta, \\ y=s+t \sin \beta,\end{array}\right.\right.$ (1 as a parameter),

Substitute into the equation of $C$ to get $3(4+t \cos \alpha)^{2}+4(s+t \sin \alpha)^{2}=12$
$$
\begin{array}{l}
\Rightarrow\left(3 \cos ^{2} \alpha+4 \sin ^{2} \alpha\right) t^{2}+(24 \cos \alpha+8 s \sin \alpha) t+4 s^{2}+36=0 \\
\Rightarrow|M A| \cdot|M B|=\left|t_{A} t_{B}\right|=\frac{4 s^{2}+36}{3 \cos ^{2} \alpha+4 \sin ^{2} \alpha},
\end{array}
$$

Similarly, we can get $|M P| \cdot|M Q|=\frac{4 s^{2}+36}{3 \cos ^{2} \beta+4 \sin ^{2} \beta}$. Therefore,
$$
3 \cos ^{2} \alpha+4 \sin ^{2} \alpha=3 \cos ^{2} \beta+4 \sin ^{2} \beta \Rightarrow \sin ^{2} \alpha=\sin ^{2} \beta \Rightarrow \sin \alpha=-\sin \beta \Rightarrow \tan \alpha+\tan \beta=0 \text {. }
$$

Thus, the sum of the slopes of line $A B$ and line $P Q$ is 0.