5. For different real numbers $m$, the equation $y^{2}-6 m y-4 x+$ $9 m^{2}+4 m=0$ represents different parabolas. A line intersects all these parabolas, and the length of the chord intercepted by each parabola is $\frac{8 \sqrt{5}}{9}$. Then the equation of this line is $\qquad$ .
5. $y=3 x-\frac{1}{3}$.

The parabola $(y-3 m)^{2}=4(x-m)$. The vertex $(m, 3 m)$ lies on the line $y=3 x$, so the required line must be parallel to it, and can be set as $y=3 x+b$. Take any parabola $y^{2}=4 x$ (when $m=0$), substitute $y=3 x+b$ into it to get $x^{2}+\frac{6 b-4}{9} x+\frac{b^{2}}{9}=0$. We can find that
$$
\left(x_{1}-x_{2}\right)^{2}=\frac{16-48 b}{81},\left(y_{1}-y_{3}\right)^{2}=\frac{16-48 b}{9} .
$$

From $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}=\frac{8 \sqrt{5}}{9}$ we get $b=$ $-\frac{1}{3}$. Therefore, $y=3 x-\frac{1}{3}$.