10. Solution: Let point $M(3 \cos \alpha, 2 \sin \alpha)$, the points of tangency be $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and the equations of the tangent lines $M A, M B$ be
$$
\begin{aligned}
\quad x_{1} \cdot 3 \cos \alpha + y_{1} \cdot 2 \sin \alpha = 4, \quad x_{2} \cdot 3 \cos \alpha + y_{2} \\
2 \sin \alpha = 4,
\end{aligned}
$$

Thus, the equation of line $A B$ is $x \cdot 3 \cos \alpha + y \cdot 2 \sin \alpha = 4$. The points $P\left(\frac{4}{3 \cos \alpha}, 0\right)$ and $Q\left(0, \frac{2}{\sin \alpha}\right)$ are given. Therefore, 
$$
|P Q| = \sqrt{\frac{16}{9 \cos ^{2} \alpha} + \frac{4}{\sin ^{2} \alpha}} = \sqrt{\frac{52}{9} + \frac{16}{9} \tan ^{2} \alpha + 4 \cot ^{2} \alpha} \geqslant \frac{10}{3},
$$

Thus, the minimum value of $|P Q|$ is $\frac{10}{3}$. At this point, $M\left(\frac{3 \sqrt{10}}{5}, \frac{2 \sqrt{15}}{5}\right)$,
$$
\begin{array}{l}
\left(\frac{3 \sqrt{10}}{5}, -\frac{2 \sqrt{15}}{5}\right), \left(-\frac{3 \sqrt{10}}{5}, \frac{2 \sqrt{15}}{5}\right), \\
\left(-\frac{3 \sqrt{10}}{5}, -\frac{2 \sqrt{15}}{5}\right).
\end{array}
$$
10. (15 points) From a point $M$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two tangents are drawn to the circle with the minor axis as its diameter, with points of tangency $A$ and $B$. The line $AB$ intersects the $x$-axis and $y$-axis at points $P$ and $Q$, respectively. Find the minimum value of $|PQ|$.