9. (16 points) Let the ellipse $C: \frac{(x-1)^{2}}{16}+\frac{(y-1)^{2}}{9}=1$, and the line $l: y=a x+b$. If two tangents are drawn from any two points $M, N$ on the line $l$ to the ellipse $C$, and the lines connecting the points of tangency are $m, n$, then $m \parallel n$. Try to find the locus of the point $(a, b)$.

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9. Let \( M\left(x_{0}, y_{0}\right) \) be any point on the line \( l: y=a x+b \).

From the ellipse \( C: 9 x^{2}+16 y^{2}-18 x-32 y-119=0 \), the chord of contact corresponding to \( M \) is
\[
9 x_{0} x+16 y_{0} y-9\left(x+x_{0}\right)-16\left(y+y_{0}\right)-119=0 \text {. }
\]

Substituting \( y_{0}=a x_{0}+b \) into the above equation and simplifying, we get
\[
\begin{array}{l}
x_{0}(9 x+16 a y-9-16 a)- \\
{[9 x+16 y(1-b)+16 b+119]=0 .}
\end{array}
\]

Solving the system of equations
\[
\left\{\begin{array}{l}
9 x+16 a y-9-16 a=0, \\
9 x+16 y(1-b)+16 b+119=0 .
\end{array}\right.
\]

If the system of equations (2) has a solution \( G\left(x_{1}, y_{1}\right) \), then for any point \( M \) on \( l \), the corresponding chord of contact will pass through point \( G \), which contradicts the given condition.
Therefore, the system of equations (2) has no solution, i.e.,
\[
\frac{9}{9}=\frac{16 a}{16(1-b)} \neq \frac{-9-16 a}{16 b+119}, a+b=1 \text {. }
\]

Substituting into equation (1), we get
\[
\left(x_{0}-1\right)(9 x+16 a y-9-16 a)-144=0 \text {, }
\]

which is the chord of contact corresponding to the point \( M\left(x_{0}, y_{0}\right) \) \(\left(x_{0} \neq 1\right)\).
The point \((1,1)\) lies on the line \( l \) and is inside the ellipse, so it has no corresponding chord of contact.
Thus, the required locus of points is the line \( a+b=1 \).