17. (18 points) Given the ellipse $C$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {, }
$$

passing through the point $P\left(3, \frac{16}{5}\right)$, with an eccentricity of $\frac{3}{5}$. A line $l$ with slope $k$ is drawn through the right focus of the ellipse $C$, intersecting the ellipse $C$ at points $A$ and $B$. The slopes of $P A$ and $P B$ are denoted as $k_{1}$ and $k_{2}$, respectively.
(1) Find the standard equation of the ellipse;
(2) If $k_{1}+k_{2}=0$, find the real number $k$.
17. (1) From the given information,
$$
\begin{array}{l}
\frac{9}{a^{2}}+\frac{256}{25 b^{2}}=1, \frac{a^{2}-b^{2}}{a^{2}}=\frac{9}{25} \\
\Rightarrow a^{2}=25, b^{2}=16 .
\end{array}
$$

Thus, the equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$.
(2) It is easy to see that the coordinates of the right focus of the ellipse are $(3,0)$.
(i) When $0<k<+\infty$, the line $l: y=k(x-3)$.

Combining with the equation of the ellipse, we get
$$
\left(16+25 k^{2}\right) x^{2}-150 k^{2} x+225 k^{2}-400=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=\frac{150 k^{2}}{16+25 k^{2}}, x_{1} x_{2}=\frac{225 k^{2}-400}{16+25 k^{2}} \text {. }
$$

Also, $k_{1}=\frac{y_{1}-\frac{16}{5}}{x_{1}-3}, k_{2}=\frac{y_{2}-\frac{16}{5}}{x_{2}-3}$, and
$$
y_{1}=k\left(x_{1}-3\right), y_{2}=k\left(x_{2}-3\right) \text {, }
$$

Thus, $k_{1}+k_{2}$
$=\frac{1536-2560 k}{5\left(16+25 k^{2}\right)\left(x_{1}-3\right)\left(x_{2}-3\right)}=0$
$\Rightarrow k=\frac{3}{5}$.
(ii) When $k=0$, $k_{1}=\frac{2}{5}, k_{2}=-\frac{8}{5}$.

Then $k_{1}+k_{2}=-\frac{6}{5} \neq 0$, which does not meet the requirement.
(iii) When $k$ does not exist, $k_{1}$ and $k_{2}$ do not exist, which does not meet the requirement.
In summary, $k=\frac{3}{5}$.