Example 5: Through the right focus $F$ of the hyperbola $x^{2}-y^{2}=1$, a chord $M N$ is drawn intersecting the right branch, and $P$ is the midpoint of the chord $M N$.
(1) Find the equation of the locus of point $P$;
(2) Draw $P Q \perp M N$ intersecting the $x$-axis at point $Q$, prove that $\frac{|M N|}{|F Q|}=\sqrt{2}$.
(1) Since the chord $M N$ passes through the right focus $F$, we can consider using the polar equation of the hyperbola to solve the problem. Establish a polar coordinate system with the right focus $F(\sqrt{2}, 0)$ as the pole and $F x$ as the polar axis. Then the polar equation of the right branch of the hyperbola is
$$
\rho=\frac{1}{1-\sqrt{2} \cos \theta}(\rho \in \mathbf{R}) .
$$

Let $M\left(\rho_{1}, \theta\right) 、 N\left(\rho_{2}, \pi+\theta\right), \theta \in$ $\left[-\frac{\pi}{2},-\frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{\pi}{2}\right), P(\rho, \theta)$. Then
$$
\rho_{1}=\frac{1}{1-\sqrt{2} \cos \theta}, \rho_{2}=\frac{1}{1+\sqrt{2} \cos \theta} \text {. }
$$

Therefore, $\rho=\frac{\rho_{1}-\rho_{2}}{2}=\frac{\sqrt{2} \cos \theta}{1-2 \cos ^{2} \theta}$.
This is the polar equation of the trajectory of the midpoint $P$ of $M N$.

Next, we convert the polar equation to a Cartesian coordinate equation.
Multiplying both sides of $\rho-2 \rho \cos ^{2} \theta=\sqrt{2} \cos \theta$ by $\rho$ gives
$$
\begin{array}{l}
\rho^{2}-2 \rho^{2} \cos ^{2} \theta=\sqrt{2} \rho \cos \theta \\
\Rightarrow \rho^{2} \sin ^{2} \theta-\rho^{2} \cos ^{2} \theta=\sqrt{2} \rho \cos \theta>0 \\
\Rightarrow y^{2}-(x-\sqrt{2})^{2}=\sqrt{2}(x-\sqrt{2}) . \\
\text { Hence }\left(x-\frac{\sqrt{2}}{2}\right)^{2}-y^{2}=\frac{1}{2}(x \geqslant \sqrt{2}) .
\end{array}
$$
(2) Note that
$$
\begin{array}{l}
|M N|=\rho_{1}+\rho_{2}=\frac{1}{1-\sqrt{2} \cos \theta}+\frac{1}{1+\sqrt{2} \cos \theta} \\
=\frac{2}{1-2 \cos ^{2} \theta} .
\end{array}
$$

And $|F Q|=\frac{|F P|}{\cos \theta}=\frac{\frac{\sqrt{2} \cos \theta}{1-2 \cos ^{2} \theta}}{\cos \theta}=\frac{\sqrt{2}}{1-2 \cos ^{2} \theta}$.
Therefore, $\frac{|M N|}{|F Q|}=\frac{2}{\sqrt{2}}=\sqrt{2}$.