已知椭圆 $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ 上四点 $A,B,C,D$，$AB,CD$ 交于定点 $P(x_0,y_0)$，$AB,CD$ 的中点分别为 $M,N$，斜率分别为 $k_1,k_2$，若 $\alpha k_1k_2+\beta(k_1+k_2)+\gamma =0$，求证：$MN$ 过定点．
解析
设 $P(x_0,y_0),M(x_1,y_1),N(x_2,y_2)$，则\[\begin{cases} k_1=\frac{y_1-y_0}{x_1-x_0}=-\frac{b^2}{a^2}\cdot \frac{x_1}{y_1},\\ k_2=\frac{y_2-y_0}{x_2-x_0}=-\frac{b^2}{a^2}\cdot \frac{x_2}{y_2},\end{cases}\]于是\[\alpha\cdot \dfrac{y_1-y_0}{x_1-x_2}\cdot \left(-\dfrac{b^2}{a^2}\cdot \dfrac{x_2}{y_2}\right)+\beta\left(\dfrac{y_1-y_0}{x_1-x_2}+\left(-\dfrac{b^2}{a^2}\cdot \dfrac{x_2}{y_2}\right)\right)+\gamma =0,\]即\[(y_1-y_0)(\beta a^2y_2-\alpha b^2x_2)+(x_1-x_0)(\gamma a^2y_2-\beta b^2x_2)=0,\]同理，有\[(y_2-y_0)(\beta a^2y_1-\alpha b^2x_1)+(x_2-x_0)(\gamma a^2y_1-\beta b^2x_1)=0,\]两式相减可得\[(\alpha b^2+\gamma a^2)(x_1y_2-x_2y_1)+(\beta \cdot a^2 y_0+\gamma a^2 x_0)(y_1-y_2)+(\alpha b^2y_0+\beta b^2 x_0)(x_2-x_1)=0,\]与直线 $MN$ 的方程\[(x_1y_2-x_2y_1)+(y_1-y_2)x+y(x_2-x_1)=0\]对比系数即可得 $MN$ 过定点．