Example 8 In the Cartesian coordinate system, given $A_{1}$ $(-3,0), A_{2}(3,0), P(x, y), M\left(\sqrt{x^{2}-9}, 0\right)$. If a real number $\lambda$ makes the vectors $\overrightarrow{A_{1} P}, \lambda \overrightarrow{O M}, \overrightarrow{A_{2} P}$ satisfy $\lambda^{2} \cdot(\overrightarrow{O M})^{2}$ $=\overrightarrow{A_{1} P} \cdot \overrightarrow{A_{2} P}$,
(1) Find the equation of the trajectory of point $P$, and determine what kind of curve the trajectory of point $P$ is;
(2) When $\lambda=\frac{\sqrt{3}}{3}$, a line passing through point $A_{1}$ with a slope of 1 intersects the curve from (1) at another point $B$. Can a point $C$ be found on the line $x=-9$ such that $\triangle A_{1} B C$ is an equilateral triangle?
Solution: (1) From the given, we have $\overrightarrow{A_{1} P}=(x+3, y), \overrightarrow{A_{2} P}=$
$$
\begin{array}{l}
(x-3, y), \overrightarrow{O M}=\left(\sqrt{x^{2}-9}, 0\right) . \\
\because \lambda^{2} \cdot(\overrightarrow{O M})^{2}=\overrightarrow{A_{1} P} \cdot \overrightarrow{A_{2} P}, \\
\therefore \lambda^{2} \cdot\left(x^{2}-9\right)=x^{2}-9+y^{2},
\end{array}
$$
$\therefore$ The equation of the trajectory of point $P$ is $\left(1-\lambda^{2}\right) x^{2}+y^{2}=$ $9\left(1-\lambda^{2}\right)$,

When $1-\lambda^{2}>0$ and $\lambda \neq 0$, i.e., $\lambda \in(-1,0) \cup$ $(0,1)$, we have $\frac{x^{2}}{9}+\frac{y^{2}}{9\left(1-\lambda^{2}\right)}=1$, the trajectory of point $P$ is an ellipse;

When $\lambda=0$, the equation is $x^{2}+y^{2}=9$, the trajectory of point $P$ is a circle;

When $1-\lambda^{2}<0$, i.e., $\lambda \in(-\infty,-1) \cup(1$, $+\infty)$, the equation is $\frac{x^{2}}{9}+\frac{y^{2}}{9\left(1-\lambda^{2}\right)}=1$, the trajectory of point $P$ is a hyperbola;

When $1-\lambda^{2}=0$, i.e., $\lambda= \pm 1$, the equation is $y=0$, the trajectory of point $P$ is a straight line.
(2) The equation of the line passing through point $A_{1}$ with a slope of 1 is $y$ $=x+3$,
When $\lambda=\frac{\sqrt{3}}{3}$, the equation of the curve is $\frac{x^{2}}{9}+\frac{y^{2}}{6}=1$,
(1) Substituting (2) into (1) and rearranging gives $5 x^{2}+18 x+9=0$,
$\therefore x_{1}=-3, x_{2}=-\frac{3}{5}$.
$\therefore\left|A_{1} B\right|=\sqrt{1+k^{2}}\left|x_{1}-x_{2}\right|=\frac{12}{5} \sqrt{2}$.
Let $C(-9, y),\left|A_{1} C\right|=\sqrt{y^{2}+36}$,
$\because \triangle A_{1} B C$ is an equilateral triangle, $\left|A_{1} B\right|=\left|A_{1} C\right|$,
$\therefore \sqrt{y^{2}+36}=\frac{12}{5} \sqrt{2}$,
$\therefore y^{2}+36=\frac{288}{25}$,
$\therefore y^{2}<0$,
$\therefore$ There is no point $C$ on the line $x=-9$ that makes $\triangle A_{1} B C$ an equilateral triangle.