10. (1) Let point $P(x, y)$.

From the problem, we know $a=\sqrt{2} b$, and the ellipse $C: x^{2}+2 y^{2}=a^{2}$. Then $\overrightarrow{P P_{1}} \cdot \overrightarrow{P P_{2}}=x^{2}+y^{2}-1=-y^{2}+a^{2}-1$.
When $y= \pm b$, the above expression achieves its minimum value, i.e.,
$$
\begin{array}{l}
a^{2}-b^{2}-1=\frac{a}{2} \Rightarrow \frac{a^{2}}{2}-1=\frac{a}{2} \\
\Rightarrow a=2, b=\sqrt{2} .
\end{array}
$$

Thus, the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{2}=1$.
(2) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M\left(x_{0}, y_{0}\right)$.

From $\left\{\begin{array}{l}x^{2}+2 y^{2}=4, \\ y=k x+m\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow\left(2 k^{2}+1\right) x^{2}+4 m k x+2 m^{2}-4=0 \\
\Rightarrow x_{1}+x_{2}=-\frac{4 m k}{2 k^{2}+1}, x_{1} x_{2}=\frac{2 m^{2}-4}{2 k^{2}+1} .
\end{array}
$$

Then the distance from point $O$ to the line $l$ is $d=\frac{|m|}{\sqrt{k^{2}+1}}$.
$$
\begin{array}{l}
\text { Hence } S=\frac{1}{2} d|A B| \\
=\frac{1}{2} \cdot \frac{|m|}{\sqrt{k^{2}+1}} \sqrt{k^{2}+1} \text {. } \\
\sqrt{\left(-\frac{4 m k}{2 k^{2}+1}\right)^{2}-4 \cdot \frac{2 m^{2}-4}{2 k^{2}+1}} \\
=\sqrt{2} \frac{\sqrt{m^{2}\left(4 k^{2}+2-m^{2}\right)}}{2 k^{2}+1} \\
\leqslant \sqrt{2} \frac{\frac{m^{2}+\left(4 k^{2}+2-m^{2}\right)}{2}}{2 k^{2}+1}=\sqrt{2} . \\
\end{array}
$$

Thus, $S$ achieves its maximum value $\sqrt{2}$, if and only if
$$
m^{2}=4 k^{2}+2-m^{2} \text {, }
$$

i.e., $m^{2}=2 k^{2}+1$.
At this time, $x_{0}=\frac{x_{1}+x_{2}}{2}=-\frac{2 m k}{2 k^{2}+1}=-\frac{2 k}{m}$,
$$
y_{0}=k x_{0}+m=-\frac{2 k^{2}}{m}+m=\frac{1}{m} \text {, }
$$

i.e., $m=\frac{1}{y_{0}}, k=-\frac{m}{2} x_{0}=-\frac{x_{0}}{2 y_{0}}$.
Substituting into equation (1) and simplifying, we get
$$
\frac{x_{0}^{2}}{2}+y_{0}^{2}=1\left(y_{0} \neq 0\right) \text {, }
$$

i.e., the trajectory of point $M$ is the ellipse $C_{1}: \frac{x^{2}}{2}+y^{2}=1(y \neq 0)$, and $P_{1} 、 P_{2}$ are the left and right foci of the ellipse $C_{1}$, i.e.,
$$
\left|M P_{1}\right|+\left|M P_{2}\right|=2 \sqrt{2} \text {. }
$$

Let $t=\left|M P_{1}\right|$, then $t \in(\sqrt{2}-1, \sqrt{2}+1)$.
Thus, $T=\frac{1}{\left|M P_{1}\right|^{2}}-2\left|M P_{2}\right|$
$$
\begin{array}{l}
=\frac{1}{t^{2}}-2(2 \sqrt{2}-t) \\
=\frac{1}{t^{2}}+2 t-4 \sqrt{2} .
\end{array}
$$

Thus, $T^{\prime}=2-\frac{2}{t^{3}}$.
Let $T^{\prime} \geqslant 0$, we get $t \geqslant 1$, i.e., $T$ is monotonically decreasing on the interval $(\sqrt{2}-1,1)$ and monotonically increasing on the interval $(1, \sqrt{2}+1)$, and
$$
\begin{array}{l}
T(1)=3-4 \sqrt{2}, \\
T(\sqrt{2}-1)=1>T(\sqrt{2}+1)=5-4 \sqrt{2} .
\end{array}
$$

Thus, the range of $T$ is $[3-4 \sqrt{2}, 1)$.
10. (20 points) 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, $P$ 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) Let the line $l: y=k x+m$ intersect the ellipse $C$ at points $A$ and $B$, $O$ be the origin, and $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B})$. When the area of $\triangle A O B$ is maximized, find the range of $T=\frac{1}{\left|M P_{1}\right|^{2}}-2\left|M P_{2}\right|$.