4. Through the point $(\sqrt{5}, \sqrt{2})$ inside the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, two chords $A B$ and $C D$ are drawn. Tangents to the ellipse are drawn through $A, B$ and intersect at $E$, and tangents to the ellipse are drawn through $C, D$ and intersect at $F$. Then the equation of the line $E F$ is
4. Solution: Let $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, then it is easy to know that the equation of $A B$ is $\frac{x_{1}}{9} x+\frac{y_{1}}{5} y=1$. Furthermore, since $A B$ passes through the point $(\sqrt{5}, \sqrt{2})$, we have $\frac{x_{1}}{9} \cdot \sqrt{5}+\frac{y_{1}}{5} \cdot \sqrt{2}=1$, which means point $E$ lies on the line $\frac{\sqrt{5}}{9} x+\frac{\sqrt{2}}{5} y=1$. Similarly, point $F$ also lies on this line. By the principle that two points determine a line, the required line is $\frac{\sqrt{5}}{9} x+\frac{\sqrt{2}}{5} y=1$.