Example 1. Find the center of the hyperbola $x^{2}+2 x y-3 y^{2}+2 x-4 y$ $=0$, the diameter passing through $(1,1)$, and the conjugate diameter, and also find its asymptotes.
Let $S \equiv x_{1}^{2}+2 x_{1} x_{2}-3 x_{2}^{2}+2 x_{1} x_{3}$ $-4 x_{2} x_{3}=0$.
(1) The coordinates of the center satisfy the equations
$$
\left.\begin{array}{l}
\frac{\partial S}{\partial x_{1}}=0 \\
\frac{\partial S}{\partial x_{2}}=0
\end{array}\right\}
$$

Thus, the center is $\left(-\frac{1}{4},-\frac{3}{4}\right)$.
(2) Let the equation of the diameter be $\frac{\partial S}{\partial x_{1}}+k \frac{\partial S}{\partial x_{2}}=0$
$$
\begin{array}{l}
\Leftrightarrow\left(x_{1}+x_{2}+x_{3}\right)+k\left(x_{1}-3 x_{2}-2 x_{3}\right) \\
=0 .
\end{array}
$$

By passing through $(1,1)$, we get $k=\frac{3}{4}$. Therefore, $7 x_{1}-5 x_{2}-2 x_{3}=0 \Longleftrightarrow 7 x-5 y-2=0$ is the required equation.
(3) Let the equation of the conjugate diameter be $\frac{\partial S}{\partial x_{1}}+k \frac{\partial S}{\partial x_{2}}$ $=0$.

Where $k, \bar{k}$ satisfy $\bar{k}=-\frac{a_{11}+k a_{12}}{a_{12}+k a_{22}}$.
Thus, $k=\frac{7}{5}, 12 x-16 y-9=0$ is the required equation.
(4) Let the equation of the asymptotes be $\frac{\partial S}{\partial x_{1}}+k \frac{\partial S}{\partial x_{2}}=0$.

Where $k$ satisfies $a_{22} k^{2}+2 a_{12} k+a_{11}=0$
$$
\Rightarrow-3 k^{2}+2 k+1=0 \Rightarrow k_{1}=1, k_{2}=-\frac{1}{3} \text {. }
$$

Thus, $2 x+6 y+5=0$ and $2 x-2 y-1=0$ are the two asymptotes.