4. For a regular tetrahedron $V-ABC$ with base edge length $a$ and side edge length $b$, let $M$ be a point on the height $VO$ such that $\frac{VM}{MO}=\frac{b}{a}$. A plane passing through $M$ is parallel to the side edge $VA$ and the base edge $BC$. The area of the section formed by this plane cutting the regular tetrahedron is
4. $\frac{2}{9} \cdot \frac{a b^{2}(3 a+b)}{(a+b)^{2}}$.

Make a section $E F G H$, it is easy to know it is a rectangle. Let it intersect the base at $N$, and intersect $V D$ at $P$. Since
$$
\begin{array}{l}
A D=\frac{\sqrt{3}}{2} a, A O= \\
\frac{\sqrt{3}}{3} a, O D=\frac{\sqrt{3}}{6} a, \frac{A N}{N D} \\
=\frac{V M}{M O}=\frac{b}{a} \text {, then } \\
\hat{L N}=\frac{b}{a+b} \cdot A O \\
=\frac{\sqrt{3} a t}{3(a+b)} \text {. } \\
\text { Also } \frac{H G}{a}=\frac{A N}{A D}=\frac{2}{3} \cdot \frac{b}{a+b} \text {, } \\
\therefore H G=\frac{2}{3} \cdot \frac{a b}{a+b} \text {. } \\
\text { Also } \frac{N P}{b}=\frac{N D}{A D}=\frac{\frac{\sqrt{3}}{2} a-A N}{\frac{\sqrt{3}}{2} a}=1-\frac{2}{3} \cdot \frac{b}{a+b} \text {, } \\
\therefore N P=b-\frac{2}{3} \cdot \frac{b^{2}}{a+b} \text {. } \\
S_{\text {section }}=H G \cdot N P=\frac{2}{3} \cdot \frac{a b}{a+b}\left(b-\frac{2 b^{2}}{3(a+b)}\right) \\
=\frac{2}{9} \cdot \frac{a b^{2}(3 a+b)}{(a+b)^{2}} \text {. } \\
\end{array}
$$