已知椭圆 $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ 的内接三角形 $ABC$ 的外心为 $P$，$O$ 为坐标原点，直线 $AB,BC,CA,OD$ 的斜率均存在，求证：直线 $AB,BC,CA,OP$ 的斜率之积为定值．
解析
设 $AB,AC$ 的中点分别为 $M(x_1,y_1),N(x_2,y_2)$，则根据椭圆的垂径定理，有\[ k_{AB}\cdot k_{OM}=k_{AC}\cdot k_{ON}=-\dfrac{b^2}{a^2},\]进而\[k_{AB}=-\dfrac{b^2x_1}{a^2y_1},\quad k_{AC}=-\dfrac{b^2x_2}{a^2y_2},\quad k_{BC}=\dfrac{y_1-y_2}{x_1-x_2},\]设 $P(x_0,y_0)$，根据圆的垂径定理，有\[k_{AB}\cdot k_{PM}=k_{AC}\cdot k_{PN}=-1\implies -\dfrac{b^2x_1}{a^2y_1}\cdot \dfrac{y_1-y_0}{x_1-x_0}=-\dfrac{b^2x_2}{a^2y_2}\cdot \dfrac{y_2-y_0}{x_2-x_0}=-1,\]于是\[\begin{cases} (a^2-b^2)x_1y_1=a^2x_0y_1-b^2x_1y_0,\\ (a^2-b^2)x_2y_2=a^2x_0y_2-b^2x_2y_0,\end{cases}\implies \dfrac{x_1y_1}{x_2y_2}=\dfrac{a^2x_0y_1-b^2x_1y_0}{a^2x_0y_2-b^2x_2y_0},\]于是由合分比定理可得\[\dfrac{a^2x_0(x_2y_1-x_1y_2)}{-b^2y_0(x_1y_2-x_2y_1)}=\dfrac{x_1x_2y_1-x_1x_2y_2}{x_1y_1y_2-x_2y_1y_2}\iff \dfrac{a^2x_0}{b^2y_0}=\dfrac{x_1x_2(y_1-y_2)}{y_1y_2(x_1-x_2)},\]因此\[\dfrac{a^2}{b^2}=k_{OP}\cdot k_{MN}\cdot \left(-\dfrac{a^2}{b^2}k_{AB}\right)\cdot \left(-\dfrac{a^2}{b^2}k_{AC}\right)\iff k_{AB}\cdot k_{AC}\cdot k_{BC}\cdot k_{OG}=\dfrac{b^2}{a^2},\]为定值．
备注 在双曲线中，对应斜率之积为 $-\dfrac{b^2}{a^2}$．
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