6. If the parabola $y=-\frac{x^{2}}{4}+4$ intersects the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{3^{2}}$ $=1$ at four distinct points, then the value of the positive number $a$ is
6. $a>\sqrt{7}+1$.

Combining the equations $y=-\frac{x^{2}}{4}+4$ and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{3^{2}}=1$, and eliminating $x$, we obtain the quadratic equation in $y$:
$$
a^{2} y^{2}-36 y+\left(144-9 a^{2}\right)=0.
$$

Let $f(y)=a^{2} y^{2}-36 y+\left(144-9 a^{2}\right)$, which is an upward-opening parabola. The necessary and sufficient condition for equation (1) to have two roots is
$$
\left\{\begin{array}{l}
-36, \\
9 a^{4}-144 a^{2}+18^{2}>0.
\end{array}\right.
$$

Solving this, we get $a^{2}>8+2 \sqrt{7}$.
Since $a$ is positive, taking the square root gives $a>\sqrt{7}+1$. Note: The two roots of equation (1) must lie within $(-3,3)$.