10. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ intersects the line $x=-\sqrt{2} b$ at exactly one point, point $P$ is any point on the ellipse $C$, $P_{1}(-1,0), P_{2}(1,0)$, and the minimum value of $\overrightarrow{P P}_{1} \cdot \overrightarrow{P P}_{2}$ is $\frac{a}{2}$.
(1) Find the equation of the ellipse $C$;
(2) Suppose the line $l: y=k x+m$ intersects the ellipse $C$ at two distinct points $A, B$, and point $O$ is the origin, and $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B})$. When the area $S$ of $\triangle A O B$ is maximized, find the range of $T=\frac{1}{\left|M P_{1}\right|^{2}}-2\left|M P_{2}\right|$.
Solution: (1) Let point $P(x, y)$, from the problem we know $a=\sqrt{2} b, C: x^{2}+2 y^{2}=a^{2}$, then
$$
\overrightarrow{P P_{1}} \cdot \overrightarrow{P P_{2}}=x^{2}+y^{2}-1=-y^{2}+a^{2}-1
$$

When $y= \pm b$, $\overrightarrow{P P_{1}} \cdot \overrightarrow{P P_{2}}$ reaches its minimum value, i.e., $a^{2}-b^{2}-1=\frac{a}{2}$, $\Rightarrow \frac{a^{2}}{2}-1=\frac{a}{2} \Rightarrow a=2, b=\sqrt{2}$, hence the ellipse $C$ is $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$; $\qquad$
(2) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M\left(x_{0}, y_{0}\right)$, then

From $\left\{\begin{array}{l}x^{2}+2 y^{2}=4 \\ y=k x+m\end{array}\right.$ we get $\left(2 k^{2}+1\right) x^{2}+4 m k x+2 m^{2}-4=0$
$\Rightarrow x_{1}+x_{2}=-\frac{4 m k}{2 k^{2}+1}, x_{1} x_{2}=\frac{2 m^{2}-4}{2 k^{2}+1}$, the distance from point $O$ to line $l$ is $d=\frac{|m|}{\sqrt{k^{2}+1}}$,
$S=\frac{1}{2} \cdot d \cdot|A B|=\frac{1}{2} \cdot \frac{|m|}{\sqrt{k^{2}+1}} \cdot \sqrt{k^{2}+1} \cdot \sqrt{\left(-\frac{4 m k}{2 k^{2}+1}\right)^{2}-4 \cdot \frac{2 m^{2}-4}{2 k^{2}+1}}$
$$
=\sqrt{2} \cdot \frac{\sqrt{m^{2}\left(4 k^{2}+2-m^{2}\right)}}{2 k^{2}+1} \leq \sqrt{2} \cdot \frac{\frac{m^{2}+\left(4 k^{2}+2-m^{2}\right)}{2}}{2 k^{2}+1}=\sqrt{2}
$$
$S$ reaches its maximum value $\sqrt{2}$, if and only if $m^{2}=4 k^{2}+2-m^{2}$, i.e., $m^{2}=2 k^{2}+1$, (1) .............10 points
At this time, $x_{0}=\frac{x_{1}+x_{2}}{2}=-\frac{2 m k}{2 k^{2}+1}=-\frac{2 k}{m}, y_{0}=k x_{0}+m=-\frac{2 k^{2}}{m}+m=\frac{1}{m}$, i.e., $m=\frac{1}{y_{0}}, k=-\frac{m}{2} x_{0}=-\frac{x_{0}}{2 y_{0}}$ substituting into (1) and rearranging gives $\frac{x_{0}{ }^{2}}{2}+y_{0}{ }^{2}=1\left(y_{0} \neq 0\right)$,

i.e., the trajectory of point $M$ is the ellipse $C_{1}: \frac{x^{2}}{2}+y^{2}=1(y \neq 0)$
and points $P_{1}, P_{2}$ are the left and right foci of ellipse $C_{1}$, i.e., $\left|M P_{1}\right|+\left|M P_{2}\right|=2 \sqrt{2} \ldots \ldots \ldots \ldots .15$ points. Let $t=\left|M P_{1}\right|$, then $t \in(\sqrt{2}-1, \sqrt{2}+1)$
Thus $T=\frac{1}{\left|M P_{1}\right|^{2}}-2\left|M P_{2}\right|=\frac{1}{t^{2}}-2(2 \sqrt{2}-t)=\frac{1}{t^{2}}+2 t-4 \sqrt{2}$, then $T^{\prime}=2-\frac{2}{t^{3}}$, let $T^{\prime} \geq 0$ we get $t \geq 1$, i.e., $T$ is monotonically decreasing in $(\sqrt{2}-1,1)$ and monotonically increasing in $(1, \sqrt{2}+1)$, and $T(1)=3-4 \sqrt{2}, T(\sqrt{2}-1)=1>T(\sqrt{2}+1)=5-4 \sqrt{2}$, hence the range of $T$ is $[3-4 \sqrt{2}, 1) \ldots \ldots \ldots \ldots . . . . . . .$. points