10. 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.
It is known that points $A(1,1), B(1,-1)$ are on the ellipse.
5 points
Let the center of symmetry of the ellipse be $(a, b)$. Then, by symmetry, points $A^{\prime}(2 a-1,2 b-1), B^{\prime}(2 a-1,2 b+1)$ are also on the ellipse. Substituting into the equation, we get:
$$
\begin{array}{l}
17(2 a-1)^{2}-16(2 a-1)(2 b-1)+4(2 b-1)^{2}-34(2 a-1)+16(2 b-1)+13=0 \\
17(2 a-1)^{2}-16(2 a-1)(2 b+1)+4(2 b+1)^{2}-34(2 a-1)+16(2 b+1)+13=0
\end{array}
$$
Subtracting (2) from (1) and simplifying, we get:
$$
b=2 a-2
$$

Substituting (3) into (1) and simplifying, we get: $4 a^{2}-8 a+4=0$, solving for $a=1$, thus $b=0$, so the center of symmetry of the ellipse is $(1,0)$.
10 points
Since the axis of symmetry passes through the center of symmetry, we can set the equation of the axis of symmetry as $y=k(x-1)(k \neq 0)$.
Let the symmetric point of $A(1,1)$ about the line $y=k(x-1)$ be $A_{0}\left(x_{0}, y_{0}\right)$, then we have
$$
\left\{\begin{array} { l } 
{ \frac { y _ { 0 } + 1 } { 2 } = k ( \frac { x _ { 0 } + 1 } { 2 } - 1 ) } \\
{ \frac { y _ { 0 } - 1 } { x _ { 0 } - 1 } = - \frac { 1 } { k } }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{0}=\frac{k^{2}+2 k+1}{k^{2}+1} \\
y_{0}=\frac{k^{2}-1}{k^{2}+1}
\end{array}\right.\right.
$$
15 points
Since $A_{0}$ is on the ellipse, we have
$$
17 x_{0}^{2}-16 x_{0} y_{0}+4 y_{0}^{2}-34 x_{0}+16 y_{0}+13=0
$$

Substituting (4) into the above equation and simplifying, we get $k\left(8 k^{2}-13 k-8\right)=0$.
Since $k \neq 0$, then $8 k^{2}-13 k-8=0$, solving for $k=\frac{13 \pm 5 \sqrt{17}}{16}$, so the equation of the axis of symmetry of the ellipse is
$$
y=\frac{13 \pm 5 \sqrt{17}}{16}(x-1)
$$
20 points