4. Given that $P(-2,3)$ is a point on the graph of the inverse proportion function $y=\frac{k}{x}$, $Q$ is a moving point on the branch of the hyperbola in the fourth quadrant. A line is drawn through point $Q$ such that it intersects the hyperbola $y=\frac{k}{x}$ at only one point, and intersects the $x$-axis and $y$-axis at points $C$ and $D$, respectively. Another line $y=\frac{3}{2} x+6$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then the minimum value of the area of quadrilateral $A B C D$ is $\qquad$.
4. 48.

It is known that $k=-6, y=-\frac{6}{x}, A(-4,0), B(0,6)$.
Let the line passing through point $Q$ be $y=a x+b$.
Then $\left\{\begin{array}{l}y=a x+b, \\ y=-\frac{6}{x}\end{array}\right.$ has only one solution.
Eliminating and rearranging gives $a x^{2}+b x+6=0$.
By $\Delta=b^{2}-24 a=0 \Rightarrow b^{2}=24 a$.
Then $O C \cdot O D=(-b)\left(-\frac{b}{a}\right)=24=O A \cdot O B$.
Let $O C=m$. Then
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=3 m+\frac{48}{m}+24 \\
=3\left(\sqrt{m}-\frac{4}{\sqrt{m}}\right)^{2}+48 .
\end{array}
$$

Obviously, when $\sqrt{m}=\frac{4}{\sqrt{m}} \Rightarrow m=4$, the minimum value 48 is obtained.