51 Given a regular quadrilateral prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, with the base edge length being $a$, and the side edge length being $b, P$ is a moving point on the diagonal $A C_{1}$, let the angle between $P C$ and the plane $A B C D$ be $\alpha$, and the angle between $P D$ and the plane $C D D_{1} C_{1}$ be $\beta$. Find the value of $\tan \alpha \cdot \tan \beta$.
51 As shown in Figure 27, since $O C_{1} \perp$ plane $A B C D$, we have $\alpha=\angle A C P$. Similarly, $\beta=\angle C_{1} D P$. Let $A P=x$. In the right triangle $A C C_{1}$,
$$
\begin{aligned}
\tan \alpha & =\frac{\sin \alpha}{\cos \beta}=\frac{x \sin A}{C P} \cdot \frac{C P}{\left(A C_{1}-x\right) \sin C_{1}} \\
& =\frac{O C_{1} \cdot x}{A C \cdot\left(A C_{1}-x\right)}
\end{aligned}
$$

Similarly,
$$
\tan \beta=\frac{A D \cdot\left(-x+A C_{1}\right)}{x \cdot C D}
$$

Therefore,
$$
\begin{aligned}
\tan \alpha \cdot \tan \beta & =\frac{O C_{1} \cdot x}{A C \cdot\left(A C_{1}-x\right)} \cdot \frac{A D \cdot\left(A C_{1}-x\right)}{x \cdot C D}=\frac{O C_{1} \cdot A D}{A C \cdot C D} \\
& =\frac{b \cdot a}{\sqrt{2} a \cdot \sqrt{a^{2}+b^{2}}}=\frac{\sqrt{2} b \cdot \sqrt{a^{2}+b^{2}}}{2\left(a^{2}+b^{2}\right)} .
\end{aligned}
$$