6. Given real numbers $x, y$ satisfy $3 x^{2}+4 y^{2}=48$. Then
$$
\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5}
$$

The maximum value is $\qquad$.
$$
6.8+\sqrt{13} \text {. }
$$

Let point $P(x, y)$ be on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and $A(2,0)$ be the right focus of the ellipse. Point $B(1,-2)$ is inside the ellipse.

Let the left focus of the ellipse be $F_{1}(-2,0)$, and the length of the major axis be $2a$. Then
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5} \\
=\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x-1)^{2}+(y+2)^{2}} \\
=|P A|+|P B|=2 a+\left(|P B|-\left|P F_{1}\right|\right) \\
\leqslant 2 a+\left|F_{1} B\right|=8+\sqrt{13} .
\end{array}
$$

The equality holds if and only if $P$ is the intersection point of the ray $B F_{1}$ with the ellipse.