Example 5 Let 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 $\qquad$ [3]
Notice that,
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}-4 x+4}+\sqrt{x^{2}+y^{2}-2 x+4 y+5} \\
=\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x-1)^{2}+(y+2)^{2}},
\end{array}
$$

and the point $P(x, y)$ lies on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$.
As shown in Figure 5, let the left and right foci of the ellipse be $F_{1}(-2, 0)$ and $F_{2}(2,0)$, with the major axis length being $2a$. Take a point $B(1,-2)$ inside the ellipse.
Then the expression we seek is
$$
\begin{array}{l}
=\left|P F_{2}\right|+|P B| \\
=2 c-\left|P F_{1}\right|+|P B| \\
\leqslant 2 c+\left|B F_{1}\right| \\
=8+\sqrt{13},
\end{array}
$$

with equality holding if and only if $P$ is the intersection point of the ray $B F_{1}$ with the ellipse, i.e.,
$$
\left\{\begin{array}{l}
\frac{x^{2}}{16}+\frac{y^{2}}{12}=1, \\
\frac{y}{x+2}=\frac{0-(-2)}{-2-1}, \\
x < -2
\end{array} \Rightarrow \left\{\begin{array}{l}
x=\frac{-32-36 \sqrt{13}}{43}, \\
y=\frac{24 \sqrt{13}-36}{43}
\end{array},\right.\right.
$$

when the equality in (1) holds.