已知二次曲线 $\Gamma:g(x,y)=0$ 上四点 $P_1,P_2,P_3,P_4$，若直线 $P_1P_3$ 与 $P_2P_4$ 交于点 $Q(x_0,y_0)$，直线 $P_1P_2$ 的直线方程为 $f(x,y)=0$，求直线 $P_3P_4$ 的方程．
解析 设 $\overrightarrow{P_3P_1}=\lambda \overrightarrow{P_1Q}$，则\[P_1=\dfrac{P_3+\lambda \cdot Q}{1+\lambda},\]于是\[ g(P_1)=g(P_3)=0\implies (1+\lambda)^2\cdot g\left(\dfrac{P_3+\lambda \cdot Q}{1+\lambda}\right)-g(P_3)=0,\]即\[2\lambda \cdot g_Q(P_3)+\lambda^2\cdot g(Q)=0\implies \lambda =-\dfrac{2\cdot g_Q(P_3)}{g(Q)}.\]又 $f(P_1)=0$，于是\[f\left(\dfrac{P_3+\lambda\cdot Q}{1+\lambda}\right)=0\iff f(P_3)+\lambda\cdot f(Q)=0,\]将 $\lambda$ 代入可得\[f(P_3)-\dfrac{2\cdot g_Q(P_3)}{g(Q)}\cdot f(Q)=0\iff g(Q)\cdot f(P_3)-2\cdot f(Q)\cdot g_Q(P_3)=0 ,\]从而直线 $P_3P_4$ 的方程为\[g(x_0)\cdot f(x,y)-2\cdot f(x_0,y_0)\cdot g_Q(x,y)=0.\]