Example 1. Simplify the quadratic curve equation:
$$
\begin{array}{l}
9 x^{2}-24 x y+16 y^{2}-20 x-14 y \\
+200=0,
\end{array}
$$

and find its focus coordinates and directrix equation.
Solve $\Delta=B^{2}-4 A C=0$, the curve is a non-central quadratic curve. From (2.4), the equation of the axis of symmetry of the curve (2.5) can be obtained as $l_{1}: 3 x-4 y+10=0$.

Substituting $l_{1}$ into (2.5) yields the intersection points of $l_{1}$ and the curve as $O\left(\frac{-2}{5}, \frac{11}{5}\right)$. The equation of the line passing through $O$ and perpendicular to $l_{1}$ is $l_{2}: 4 x+3 y-5=0$.

Based on the normal directions of $l_{1}$ and $l_{2}$, determine the positive directions of the coordinate axes, with the $x^{\prime}$ axis rotating counterclockwise to the $y^{\prime}$ axis. Therefore, $l_{1}$ is the $x^{\prime}$ axis, and $l_{2}$ is the $y^{\prime}$ axis. In the coordinate system $x^{\prime} o y^{\prime}$, the coordinates of point $P^{\prime}\left(x^{\prime}, y^{\prime}\right)$ are the deviations from the $x^{\prime}$ axis and the $y^{\prime}$ axis:
$$
\left\{\begin{aligned}
x^{\prime} & =\frac{4 x+3 y-5}{5}, \\
y^{\prime} & =\frac{3 x-4 y+10}{5} \\
x & =\frac{4 x^{\prime}-3 y^{\prime}-2}{5}, \\
y= & \frac{3 x^{\prime}+4 y^{\prime}+11}{5}
\end{aligned}\right.
$$

Substituting (2.7) into (2.5) and simplifying, we get
$$
y^{\prime 2}=4 x^{\prime} \text {. }
$$

In the coordinate system $x^{\prime} o^{\prime} y^{\prime}$, the coordinates of the focus are $(1,0)$, and in the original coordinate system $x o y$, the coordinates of the focus are $\vec{F}\left(\frac{2}{5}, \frac{14}{5}\right)$, and the equation of the directrix is $4 x+3 y=0$.

From the above example, it can be seen that using (2.4) to simplify non-central quadratic curves is simple and easy to understand, avoiding the complex and difficult-to-remember rotation formulas. It is worth noting that (2.4) can directly produce the image of the non-central quadratic curve in the original coordinate system.