(20 points) In the Cartesian coordinate system, given points $A(-2,0), B(2,0), P$ is a moving point on the plane, let $|P A|=a,|P B|=b$, satisfying $a b=2$.
(1) Find the range of values for $a$;
(2) Find the locus of point $P$, and draw its graph.
(1) Since $|A B|=4, a b=2$, by the properties of a triangle, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b \geqslant 4, \\
a+4 \geqslant b, \\
b+4 \geqslant a
\end{array}, \Rightarrow\left\{\begin{array}{l}
a+\frac{2}{a} \geqslant 4, \\
a+4 \geqslant \frac{2}{a}, \\
\frac{2}{a}+4 \geqslant a
\end{array}\right.\right. \\
\Rightarrow\left\{\begin{array}{l}
0<a \leqslant 2-\sqrt{2} \text { or } a \geqslant 2+\sqrt{2}, \\
a \geqslant \sqrt{6}-2, \\
0<a \leqslant 2+\sqrt{6} .
\end{array}\right.
\end{array}
$$

Therefore, the range of $a$ is
$$
\sqrt{6}-2 \leqslant a \leqslant 2-\sqrt{2} \text { or } 2+\sqrt{2} \leqslant a \leqslant 2+\sqrt{6} \text {. }
$$
(2) From $a b=2$, we get
$$
\sqrt{(x+2)^{2}+y^{2}} \cdot \sqrt{(x-2)^{2}+y^{2}}=2 \text {. }
$$

Thus, the equation of the trajectory of the moving point $P$ is
$$
\left[(x+2)^{2}+y^{2}\right]\left[(x-2)^{2}+y^{2}\right]=4 \text {. }
$$

From equation (1), we have
$$
y^{4}+2\left(x^{2}+4\right) y^{2}+x^{4}-8 x^{2}+12=0 \text {. }
$$

Then, $y^{2}=-\left(x^{2}+4\right)+2 \sqrt{4 x^{2}+1}$.
From $-\left(x^{2}+4\right)+2 \sqrt{4 x^{2}+1} \geqslant 0$, we get
$-\sqrt{6} \leqslant x \leqslant-\sqrt{2}$ or $\sqrt{2} \leqslant x \leqslant \sqrt{6}$.
Therefore, $y= \pm \sqrt{-\left(x^{2}+4\right)+2 \sqrt{4 x^{2}+1}}$, where $x \in[-\sqrt{6},-\sqrt{2}] \cup[\sqrt{2}, \sqrt{6}]$.

Using a graphing calculator, its curve is shown in Figure 2 (approximate graph):