10. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x C y$, the tangent line $l$ of a circle with center at the origin $C$ and radius 1 intersects the $x$-axis at point $N$ and the $y$-axis at point $M$. Point $A(3,4)$, and $\overrightarrow{A C}=$
$x \overrightarrow{A M}+y \overrightarrow{A N}$. Let
point $P(x, y)$.
(1) Find the equation of the trajectory of point $P$;
(2) Find the minimum value of $9 x^{2}+16 y^{2}$.
10. (1) Let point $M(0, m), N(n, 0)$.

From $\overrightarrow{A C}=x \overrightarrow{A M}+y \overrightarrow{A N}$
$\Rightarrow\left\{\begin{array}{l}3 x+(3-n) y=3, \\ (4-m) x+4 y=4\end{array}\right.$
$\Rightarrow\left\{\begin{array}{l}m=\frac{4(x+y-1)}{x}, \\ n=\frac{3(x+y-1)}{y}\end{array}\right.$,
$\Rightarrow 1=\frac{1}{m^{2}}+\frac{1}{n^{2}}$
$=\left(\frac{x}{4(x+y-1)}\right)^{2}+\left(\frac{y}{3(x+y-1)}\right)^{2}$
$\Rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{9}=(x+y-1)^{2}$.
Since $|m|>1$ and $|n|>1$, we have $x \neq 0$ and $y \neq 0$.
Thus, the equation of the trajectory of point $P$ is
$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}=(x+y-1)^{2}(x \neq 0 \text { and } y \neq 0) \text {. }
$$
(2) Since $x, y$ satisfy
$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}=(x+y-1)^{2} \text {, }
$$

we have $9 x^{2}+16 y^{2}=144\left(\frac{x^{2}}{16}+\frac{y^{2}}{9}\right)=144(x+y-1)^{2}$.
Let $z=x+y$. Then
$$
\frac{x^{2}}{16}+\frac{(z-x)^{2}}{9}=(z-1)^{2} \text {. }
$$

Thus, $25 x^{2}-32 z x+16 z^{2}-144(z-1)^{2}=0$ has a solution, i.e., $\Delta=(-32 z)^{2}-4 \times 25\left(16 z^{2}-144(z-1)^{2}\right) \geqslant 0$
$\Rightarrow z \geqslant \frac{5}{4}$ or $z \leqslant \frac{5}{6}$.
Therefore, when $x+y=\frac{5}{6}$, i.e., $x=\frac{8}{15}, y=\frac{3}{10}$, $9 x^{2}+16 y^{2}=144(x+y-1)^{2}$ achieves its minimum value 4.