3. As shown in Figure $1, P$ is any point on the right branch of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>$ $0, b>0 ＼mathrm{~ right ~ branch ~}$, a line passing through point $P$ intersects the two asymptotes of the hyperbola at points $P_{1}$ and $P_{2}$, respectively, and point $P$ internally divides the line segment $P_{1} P_{2}$. $O$ is the origin of the coordinate system, and $c$ is the semi-focal distance of the hyperbola. Then the minimum value of the area of $\triangle P_{1} O P_{2}$ is ( ).
(A) $a b$
(B) $a c$
(C) $b c$
(D) $\frac{a b^{2}}{c}$
3. A.

Let point $P$ divide $P_{1} P_{2}$ in the ratio $\lambda(\lambda>0), P(x, y)$, $P_{1}\left(x_{1}, \frac{b}{a} x_{1}\right), P_{2}\left(x_{2},-\frac{b}{a} x_{2}\right)$. Then
$$
\left\{\begin{array}{l}
x=\frac{x_{1}+\lambda x_{2}}{1+\lambda}, \\
y=\frac{\frac{b}{a} x_{1}-\lambda \cdot \frac{b}{a} x_{2}}{1+\lambda},
\end{array}\right.
$$

i.e., $\left\{\begin{array}{l}x=\frac{x_{1}+\lambda x_{2}}{1+\lambda}, \\ \frac{a}{b} y=\frac{x_{1}-\lambda x_{2}}{1+\lambda} .\end{array}\right.$
(1) $)^{2}$-(2) $)^{2}$ gives $x^{2}-\frac{a^{2}}{b^{2}} y^{2}=\frac{4 \lambda x_{1} x_{2}}{(1+\lambda)^{2}}$.
Since point $P(x, y)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we have
$$
x^{2}-\frac{a^{2}}{b^{2}} y^{2}=a^{2} \text {. }
$$

Thus, $x_{1} x_{2}=\frac{a^{2}(1+\lambda)^{2}}{4 \lambda}$.
Then $\left|O P_{1}\right| \cdot\left|O P_{2}\right|$
$$
\begin{array}{l}
=\sqrt{x_{1}^{2}+\left(\frac{b}{a} x_{1}\right)^{2}} \cdot \sqrt{x_{2}^{2}+\left(-\frac{b}{a} x_{2}\right)^{2}} \\
=\left(1+\frac{b^{2}}{a^{2}}\right)\left|x_{1} x_{2}\right|=\frac{\left(a^{2}+b^{2}\right)(1+\lambda)^{2}}{4 \lambda} .
\end{array}
$$

Also, $\sin \angle P_{1} O P_{2}=\frac{2 \times \frac{b}{a}}{1+\left(\frac{b}{a}\right)^{2}}=\frac{2 a b}{a^{2}+b^{2}}$, hence
$$
\begin{array}{l}
S_{\Delta P_{1} O P_{2}}=\frac{1}{2}\left|O P_{1}\right| \cdot\left|O P_{2}\right| \sin \angle P_{1} O P_{2} \\
=\frac{a b(1+\lambda)^{2}}{4 \lambda} \geqslant a b .
\end{array}
$$

Equality holds if and only if $\lambda=1$, i.e., $P$ is the midpoint of segment $P_{1} P_{2}$.