9. (Mathematics for Middle School, Issue 6, 2004, Olympiad Training Problem) Given the equation $17 x^{2}-16 x y+4 y^{2}-34 x+16 y+13=0$ represents an ellipse on the $x O y$ plane. Try to find its center of symmetry and axes of symmetry.
9. It is known that points $A(1,1), B(1,-1)$ are both on the curve.
Let the center of symmetry of the ellipse be $(a, b)$, then points $A^{\prime}(2 a-1,2 b-1), B^{\prime}(2 a-1,2 b+1)$ are also on the curve. Therefore, we have
$17(2 a-1)^{2}-16(2 a-1)(2 b-1)+4(2 b-1)^{2}-34(2 a-1)+16(2 b-1)+13=0$,
$17(2 a-1)^{2}-16(2 a-1)(2 b+1)+4(2 b+1)^{2}-34(2 a-1)+16(2 b+1)+13=0$.
Subtracting (2) from (1) and simplifying, we get $b=2 a-2$. Substituting $b=2 a-2$ into (1) and simplifying, we get
$4 a^{2}-8 a+4=0$.
Solving for $a$, we get $a=1$, thus, $b=0$.
Therefore, the center of symmetry is $(1,0)$.
Since the axis of symmetry passes through the center of symmetry, we can assume the equation of the axis of symmetry to be $y=k(x-1)$. Let the symmetric point of $A(1,1)$ with respect to the line $y=k(x-1)$ be $A_{0}\left(x_{0}, y_{0}\right)$, then we have
$$
\left\{\begin{array}{l}
\frac{y_{0}+1}{2}=k\left(\frac{x_{0}+1}{2}-1\right) \\
\frac{y_{0}-1}{x_{0}-1}=-\frac{1}{k}
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{0}=\frac{1+2 k+k^{2}}{1+k^{2}}, \\
y_{0}=\frac{k^{2}-1}{1+k^{2}} .
\end{array}\right.\right.
$$

Since $A_{0}$ is on the curve, we have
$$
17 x_{0}^{2}-16 x_{0} y_{0}+4 y_{0}^{2}-34 x_{0}+16 y_{0}+13=0 \text {. }
$$

Substituting (3) into the above equation and simplifying, we get
$$
k\left(8 k^{2}-13 k-8\right)=0 \text {. }
$$
$k=0$ is not a valid solution, hence
$$
8 k^{2}-13 k-8=0 \Rightarrow k=\frac{13 \pm 5 \sqrt{17}}{16} \text {. }
$$

Therefore, the equation of the axis of symmetry is $y=\frac{13 \pm 5 \sqrt{17}}{16}(x-1)$.