设 $ l $ 的方程为 $ x=my+t $ ,其中 $ k_1=dfrac{1}{m} $ , $ Aleft( x_1,y_1 right),Bleft( x_2,y_2 right) $ ，由 $$ left{begin{array}{l} x=my+t,\ dfrac{x^2}{a^2}+dfrac{y^2}{b^2}=1, \ end{array} right. $$ 消去 $ x $ 得 $$ left(a^2+b^2 m^2right)y^2 +2 b^2 m t y+b^2 left(t^2-a^2right) $$ 所以 $$ Delta=4 a^2 b^2 left(a^2+b^2 m^2-t^2right)>0,quad y_1+y_2=-dfrac{2 b^2 m t}{a^2+b^2 m^2},quad y_1y_2=frac{b^2 left(t^2-a^2right)}{a^2+b^2 m^2} $$ 所以 $$ dfrac{y_1+y_2}{y_1y_2}=-dfrac{2 m t}{t^2-a^2} $$ 设 $ triangle PAB $ 外心 $ Mleft( x_0,y_0 right) $ ,由 $ Pleft( -a,0 right),Aleft( x_1,y_1 right),Bleft( x_2,y_2 right) $ 及 $ |PM|=|AM| $ 得 $$ left( x_0+a right)^2+y_0^2=left( x_0-x_1 right)^2+left( y_0-y_1 right)^2 $$ 整理得 $$ 2aleft( a+x_1 right)x_0+2y_1y_0=x_1^2+y_1^2-a^2 $$ 由 $ dfrac{x_1^2}{a^2}+dfrac{y_1^2}{b^2}=1 $ 得 $ x_1^2=a^2-dfrac{a^2}{b^2}y_1^2 $ 代入上式得 $$ 2aleft( a+x_1 right)x_0+2y_1y_0=left( 1-dfrac{a^2}{b^2} right)y_1^2qquad (1) $$ 同理由 $ |PM|=|BM| $ 得 $$ 2aleft( a+x_2 right)x_0+2y_2y_0=left( 1-dfrac{a^2}{b^2} right)y_2^2qquad (2) $$ $ (1)times y_2^2-(2)times y_1^2 $ 整理得 $$ 2ay_2^2left( a+x_1 right)x_0+2y_2^2y_1y_0 =2ay_1^2left( a+x_2 right)x_0+2y_1^2y_2y_0 $$ 所以 $$ k_2=dfrac{y_0}{x_0}=dfrac{y_1^2left( a+x_2 right)-y_2^2left( a+x_1 right)}{y_2^2y_1-y_1^2y_2} $$ 将 $ x_1=my_1+t,x_2=my_2+1 $ 代入上式得 $$ begin{aligned} k_2&=dfrac{y_1^2left(my_2+a+t right)-y_2^2left( my_1+a+t right)}{y_2^2y_1-y_1^2y_2}\ &=dfrac{left(a+t right)left(y_1^2-y_2^2 right)+my_1y_2left( y_1-y_2 right)}{y_1y_2left( y_2-y_1 right)}\ &=left(a+t right)cdotdfrac{y_1+y_2 }{-y_1y_2}-m\ &=left(a+t right)cdotdfrac{2 m t}{t^2-a^2}-m\ &=frac{t+a}{t-a}cdot m end{aligned} $$ 因为 $ m=dfrac{1}{k_1} $ ,所以 $ k_1k_2=dfrac{t+a}{t-a} $ .