Example 6 Given that $l_{1}$ and $l_{2}$ are two perpendicular lines passing through the point $P(-\sqrt{2}, 0)$, and $l_{1}$ and $l_{2}$ each intersect the hyperbola $y^{2}-x^{2}=1$ at two points, respectively $A_{1}, B_{1}$ and $A_{2}, B_{2}$.
(1) Find the range of the slope $k_{1}$ of $l_{1}$;
(2) If point $A_{1}$ is exactly one of the vertices of the hyperbola, find the value of $\left|A_{2} B_{2}\right|$;
(3) If $\left|A_{1} B_{1}\right|=\sqrt{5}\left|A_{2} B_{2}\right|$, find the equations of $l_{1}$ and $l_{2}$.
Solution: (1) According to the problem, the slopes of $l_{1}$ and $l_{2}$ both exist. Since $l_{1}$ passes through the point $P(-\sqrt{2}, 0)$ and intersects the hyperbola at two points, we have
$$
\left\{\begin{array}{l}
y=k_{1}(x+\sqrt{2})(k \neq 0) \\
y^{2}-x^{2}=1
\end{array}\right.
$$

which has two different solutions.
Eliminating $y$ gives $\left(k_{1}^{2}-1\right) x^{2}+2 \sqrt{2} k_{1}^{2} x^{2}+2 k_{1}^{2}-1=0$.
Therefore, $\left|k_{1}\right| \neq 1$, and the discriminant of equation (12) is
$$
\triangle_{1}=\left(2 \sqrt{2} k_{1}^{2}\right)^{2}-4\left(k_{1}^{2}-1\right)\left(2 k_{1}^{2}-1\right)=4\left(3 k_{1}^{2}-1\right) .
$$

Let the slope of $l_{2}$ be $k_{2}$. Since $l_{2}$ passes through the point $P(-\sqrt{2}, 0)$ and intersects the hyperbola at two points, the system of equations
$$
\left\{\begin{array}{l}
y=k_{2}(x+\sqrt{2}) \quad(k \neq 0) \\
y^{2}-x^{2}=1
\end{array}\right.
$$

has two different solutions.
Eliminating $y$ gives $\quad\left(k_{2}^{2}-1\right) x^{2}+2 \sqrt{2} k_{2}^{2} x+2 k_{2}^{2}-1=0$.
Similarly, we have $k_{2}^{2}-1 \neq 0, \triangle_{2}=4\left(3 k_{2}^{2}-1\right)$.
Since $l_{1} \perp l_{2}$, we get $k_{1} k_{2}=-1$, so
$$
\left\{\begin{array} { l } 
{ 3 k _ { 1 } ^ { 2 } - 1 > 0 } \\
{ 3 k _ { 2 } ^ { 2 } - 1 , } \\
{ k _ { 1 } \cdot k _ { 2 } = - 1 , } \\
{ | k _ { 1 } | \neq 1 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
\frac{\sqrt{3}}{3}<\left|k_{1}\right|<\sqrt{3}, \\
\left|k_{1}\right| \neq 1 .
\end{array}\right.\right.
$$

Thus, $k_{1} \in(-\sqrt{3},-1) \cup\left(-1,-\frac{\sqrt{3}}{3}\right) \cup(1, \sqrt{3})$.
(2) The vertices of the hyperbola $y^{2}-x^{2}=1$ are $(0,1),(0,-1)$. Taking $A_{1}(0,1)$, we have $k_{1}=\frac{\sqrt{2}}{2}$, so $k_{2}=-\frac{1}{k_{1}}=-\sqrt{2}$.

Substituting $k_{2}=-\sqrt{2}$ into equation (13) gives $x^{2}+4 \sqrt{2} x+3=0$. Let the two intersection points of $l_{2}$ with the hyperbola be $A_{2}\left(x_{1}, y_{1}\right), B_{2}\left(x_{2}, y_{2}\right)$, then
$$
\left|A_{2} B_{2}\right|^{2}=3\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right] \text {. }
$$

Also, $x_{1}+x_{2}=-4 \sqrt{2}, x_{1} x_{2}=3$. Thus, we get $\left|A_{2} B_{2}\right|^{2}=60$.
Therefore, when taking $A_{1}(0,-1)$, by the symmetry of the hyperbola $y^{2}-x^{2}=1$ about the $x$-axis, we have $\left|A_{2} B_{2}\right|=2 \sqrt{15}$.
So, when $l_{1}$ passes through one of the vertices of the hyperbola, $\left|A_{2} B_{2}\right|=2 \sqrt{15}$.
(3) From equation (12), we know
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{-2 \sqrt{2} k_{1}^{2}}{k_{1}^{2}-1}, x_{1} x_{2}=\frac{2 k_{1}^{2}-1}{k_{1}^{2}-1}, \text { then } \\
\left|A_{1} B_{1}\right|^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=\frac{4\left(1+k_{1}^{2}\right)\left(3 k_{1}^{2}-1\right)}{\left(k_{1}^{2}-1\right)} .
\end{array}
$$

Similarly, from equation (13), we know
$$
\left|A_{2} B_{2}\right|^{2}=\frac{4\left(1+k_{1}^{2}\right)\left(3-k_{1}^{2}\right)}{k_{1}^{2}-1} .
$$

From $\left|A_{1} B_{1}\right|=\sqrt{5}\left|A_{2} B_{2}\right|$ and equations (15) and (16), we get
$$
\frac{4\left(1+k_{1}^{2}\right)\left(3 k_{1}^{2}-1\right)}{k_{1}^{2}-1}=5 \cdot \frac{4\left(1+k_{1}^{2}\right)\left(3-k_{1}^{2}\right)}{k_{1}^{2}-1},
$$

Solving, we get $k_{1}= \pm \sqrt{2}$.
Thus, when $k_{1}=\sqrt{2}$, $l_{1}: y=\sqrt{2}(x+\sqrt{2}), l_{2}: y=-\frac{\sqrt{2}}{2}(x+\sqrt{2})$
When $k_{2}=-\sqrt{2}$, $l_{1}: y=-\sqrt{2}(x+2), l_{2}: y=\frac{\sqrt{2}}{2}(x+$