Example 2. Simplify the quadratic curve equation
$$
\begin{aligned}
14 x^{2} & +24 x y+21 y^{2}-4 x+18 y \\
& -139=0 .
\end{aligned}
$$
Solve $\triangle=B^{2}-4 A C=-600<0$, then the equation represents an elliptical quadratic curve. By
$$
\begin{array}{l}
f_{x}^{\prime}(x, y)=14 x+12 y-2=0, \\
f_{y}^{\prime}(x, y)=12 x+21 y+S=0,
\end{array}
$$

we get the center of this curve as $O(1,-1)$. The parametric equations of the line system are:
$$
\left(\begin{array}{l}
x=1+t \cos \theta, \\
y=-1+t \sin \theta,
\end{array} \text { ( } t\right. \text { is a parameter) (3.4) }
$$

Substituting $(3.4)$ into $(3.3)$ and simplifying, we get
$$
\begin{array}{l}
\left(14 \cos ^{2} \theta+24 \sin \theta \cos \theta+21 \sin ^{2} \theta\right) t^{2} \\
-150=0 .
\end{array}
$$

Let $f(\theta)=14 \cos ^{2} \theta+24 \sin \theta \cos \theta$
$$
+21 \sin ^{2} \theta \text {, }
$$

then $f^{\prime}(\theta)=\frac{1}{2}(24 \cos 2 \theta+7 \sin 2 \theta)$.
When $f^{\prime}(\theta)=0$, $\operatorname{tg} \theta_{1}=\frac{4}{3}$,
$$
\begin{array}{l}
\operatorname{tg} \theta_{2}=-\frac{3}{4}, \\
t^{2}{ }_{\mathrm{max}}=30, t^{2}{ }_{\mathrm{m}_{1 \mathrm{n}}}=5 .
\end{array}
$$

Thus, by translating the origin of the coordinate system $x o y$ to $O^{\prime}(1,-1)$, and then rotating by $\theta=\operatorname{arctg} \frac{4}{3}$, in the new coordinate system $x^{\prime} o^{\prime} y^{\prime}$, the equation of the ellipse simplifies to:
$$
\frac{x^{\prime 2}}{5}+\frac{y^{\prime 2}}{30}=1 \text { . }
$$