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设$A=(0,0), B=\left(u_1, 0\right), C=\left(u_2, u_3\right), O=(x, y), C_1=\left(x_2, x_1\right), B_1=\left(x_4, x_3\right)$.
$$
\begin{array}{ll}
h_1=\left(y-u_3\right) x_2+\left(-x+u_2\right) x_1-u_2 y+u_3 x=0 &∵ C_1 \text { is on line } O C \\
h_2=2 u_1 x_2-u_1^2=0 &∵ C_1 A \equiv C_1 B \\
h_3=y x_4+\left(-x+u_1\right) x_3-u_1 y=0 &∵ B_1 \text { is on line } O B \\
h_4=2 u_2 x_4+2 u_3 x_3-u_3^2-u_2^2=0 &∵ B_1 A \equiv B_1 C \\
g=\left(u_1 u_3 x_2+u_1 u_2 x_1\right) x_4+\left(-u_1 u_2 x_2+u_1 u_3 x_1\right) x_3+\left(-u_1 u_3^2-u_1 u_2^2\right) x_1=0
&∵ \tan \left(B A C_1\right)=\tan \left(A C B_1\right) .
\end{array}
$$
求出 $h_1, \ldots, h_4$ 的 regular chain:
$$
\begin{aligned}
& f_1=\left(2 u_1 x-2 u_1 u_2\right) x_1+\left(2 u_1 u_2-u_1^2\right) y-2 u_1 u_3 x+u_1^2 u_3=0 \\
& f_2=2 u_1 x_2-u_1^2=0 \\
& f_3=\left(2 u_3 y+2 u_2 x-2 u_1 u_2\right) x_3+\left(-u_3^2-u_2^2+2 u_1 u_2\right) y=0 \\
& f_4=2 u_2 x_4+2 u_3 x_3-u_3^2-u_2^2=0
\end{aligned}
$$
Then $R_0$, the final remainder of $g$ with respect to $f_1, \ldots, f_4$, is
$$
R_0=u_1^2\left(u_3^2+u_2^2\right)\Bigl(a y^2+2 b x y+c u_3 x^2+\left(u_1 u_3^2-u_1 u_2^2-u_1^2 u_2\right) y+\left(2 u_1 u_2-u_1^2\right) u_3 x\Bigr)
$$
其中 $\begin{aligned}[t]a&=\left(2 u_2-u_1\right) u_3\\ b&=\left(-u_3^2+u_2^2-u_1 u_2+u_1^2\right)\\ c&=-a\end{aligned}$
从$c=-a$知$\operatorname{Tr}\pmatrix{a&b\\b&c}=0$, 所以是一条直角双曲线 |
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