10. (15 points) Given that $P$ is a point on the hyperbola $E: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and points $A$ and $B$ are on the two asymptotes of the hyperbola, with $A$ in the first quadrant and $B$ in the fourth quadrant. When $P$, $A$, and $B$ are collinear, find the range of the area of $\triangle A O B$.
10. As shown in Figure 6, given the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
$$

the points on the first and fourth quadrants of the two asymptotes are
$A(a m, b m)$,
$$
B(a n,-b n)(m>0, n>0) \text {. }
$$

Then the equation of the line $A B$ is
$$
(a m-a n)(y-b m)=(b m+b n)(x-a m) \text {, }
$$

which simplifies to $(b m+b n) x-(a m-a n) y-2 a b m n=0$. The distance from the origin to the line $A B$ (i.e., the height of $\triangle A O B$) is
$$
h=\frac{|2 a b m n|}{\sqrt{(b m+b n)^{2}+(a m-a n)^{2}}}=\frac{2 a b m n}{|A B|} .
$$

Thus, $S_{\triangle A O B}=\frac{1}{2}|A B| h=a b m n$. Furthermore, since $P, A, B$ are collinear, we have
$\left\{\begin{array}{l}x_{P}=\frac{x_{A}+\lambda x_{B}}{1+\lambda}=\frac{a(m+\lambda n)}{1+\lambda}, \\ y_{P}=\frac{y_{A}+\lambda y_{B}}{1+\lambda}=\frac{b(m-\lambda n)}{1+\lambda}\end{array}(\lambda>0)\right.$,
$\left\{\begin{array}{l}\frac{x_{P}}{a}=\frac{m+\lambda n}{1+\lambda}, \\ \frac{y_{P}}{b}=\frac{m-\lambda n}{1+\lambda} .\end{array}\right.$
Squaring and subtracting, using equation (1) (point $P$ lies on the hyperbola),
$1=\left(\frac{m+\lambda n}{1+\lambda}\right)^{2}-\left(\frac{m-\lambda n}{1+\lambda}\right)^{2}$.
Thus, $m n=\frac{(1+\lambda)^{2}}{4 \lambda}$. Substituting this into equation (2) and using the basic inequality $1+\lambda \geqslant 2 \sqrt{\lambda}$, we get
$S_{\triangle M O B}=a b \frac{(1+\lambda)^{2}}{4 \lambda} \geqslant a b$.
The minimum value of $S_{\triangle A O B}$ is $a b$ when $\lambda=1$. Furthermore, since $\lim _{\lambda \rightarrow+\infty} \frac{4 \lambda}{(1+\lambda)^{2}}=0$, the range of the area of $\triangle A O B$ is $[a b,+\infty)$.