以极点极线为背景的蝴蝶模型五种解法
【例题】已知椭圆Γ：\(\dfrac{x^2}{9}+y^2=1\)，\(A_1(-3,0)\)，\(A_2(3,0)\)，\(P\)为直线\(l：x=6\)上一动点，直线\(PA_1、PA_2\)与椭圆\(Γ\)分别交于点\(E、F\)，求证：直线\(EF\)恒过定点．
【解法1】交轨法+凑韦达
设\(EF：x=my+t，E(x_1,y_1)，F(x_2,y_2)\)，
联立\(\left\{ \begin{array}{ll} x=my+t \\ x^2+9y^2=9 \end{array} \right.\)得\((m^2+9)y^2+2mty+t^2-9=0\)
\(∴y_1+y_2=-\dfrac{2mt}{m^2+9}\)，\(y_1y_2=\dfrac{t^2-9}{m^2+9}\)
设\(P(x,y)\)，则由\(\left\{\begin{array}{ll}k_{PA_1}=k_{EA_1} \\ k_{PA_2}=k_{FA_2} \end{array} \right.\)得
\(\left\{ \begin{array}{ll} \dfrac{y}{x+3}=\dfrac{y_1}{x_1+3}=\dfrac{y_1}{my_1+t+3} \\ \dfrac{y}{x-3}=\dfrac{y_2}{x_2-3}=\dfrac{y_2}{my_2+t-3} \end{array} \right.\)
\(⇒\left\{\begin{array}{ll}\dfrac{x+3}{y}=\dfrac{my_1+t+3}{y_1} \\ \dfrac{x-3}{y}=\dfrac{my_2+t-3}{y_2} \end{array} \right.\)
\(⇒\left\{\begin{array}{ll}\dfrac{x+3}{y}=m+\dfrac{t+3}{y_1}① \\ \dfrac{x-3}{y}=m+\dfrac{t-3}{y_2}② \end{array} \right.\)
由\(①×(t-3)+②×(t+3)\)得
\(\dfrac{(t-3)(x+3)+(t+3)(x-3)}{y}=2mt+(t^2-9) \cdot \dfrac{y_1+y_2}{y_1y_2}\)
\(=2mt+(t^2-9) \cdot \dfrac{-2mt}{t^2-9}=0\)
\(⇒2tx-18=0\)
\(⇒tx=9\)
将\(x=6\)代入上式得\(t=\dfrac{3}{2}\)
故直线\(EF\)过定点\((\dfrac{3}{2},0)\)
【变式】若本题改为直线EF过定点\((\dfrac{3}{2},0)\)，则用上面的方法同样可得到\(tx=9\)，且\(t=\dfrac{3}{2}\)，从而\(x=6\)
即\(P\)点的轨迹方程为\(x=6\)
一般地，\(EF\)所过定点\((t,0)\)与点\(P\)的轨迹方程\(x=n\)满足\(tn=a^2\)
【解法2】交轨法+非对称韦达
设\(EF：x=my+t\)，\(E(x_1,y_1)\)，\(F(x_2,y_2)\)，
联立\(\left\{\begin{array}{ll}x=my+t \\ x^2+9y^2=9 \end{array} \right.\)得\((m^2+9)y^2+2mty+t^2-9=0\)
\(∴y_1+y_2=-\dfrac{2mt}{m^2+9}\)，\(y_1y_2=\dfrac{t^2)-9}{m^2+9}\)
设\(P(x,y)\)，则由\(\left\{\begin{array}{ll}k_{PA_1}=k_{EA_1} \\ k_{PA_2}=k_{FA_2} \end{array} \right.\)得
\(\left\{\begin{array}{ll}\dfrac{y}{x+3}=\dfrac{y_1}{x_1+3}① \\ \dfrac{y}{x-3}=\dfrac{y_2}{x_2-3}② \end{array} \right.\)
由\(①÷②\)得
\(\dfrac{x-3}{x+3}=\dfrac{y_1(x_2-3)}{y_2(x_1+3)}\)
\(=\dfrac{y_1(my_2+t-3)}{y_2(my_1+t+3)}\)
\(=\dfrac{my_1y_2+(t-3)y_1}{my_1y_2+(t+3)y_2}\)
\(=\dfrac{6-3}{6+3}=\dfrac{1}{3}\)
整理得\(2my_1y_2+3(t-3)y_1-(t+3)y_2=0③\)
\(∵\dfrac{y_1+y_2}{y_1y_2}=\dfrac{-2mt}{t^2-9}\)，\(∴2my_1y_2=\dfrac{t^2-9}{-t} \cdot (y_1+y_2)④\)
将\(④\)代入\(③\)得\(\dfrac{t^2-9}{-t} \cdot (y_1+y_2)+3(t-3)y_1-(t+3)y_2=0\)
整理得\((2t-3)[-(t-3)y_1+(t+3)y_2]=0\)
\(∴2t-3=0\)，即\(t=\dfrac{3}{2}\)，故直线\(EF\)恒过定点(\dfrac{3}{2},0)\)
【解法3】蝴蝶模型斜率比+曲线代换转韦达
设\(EF：x=my+t\)，\(E(x_1,y_1)\)，\(F(x_2,y_2)\)，\(P(6,n)\)，
联立\(\left\{\begin{array}{ll}x=my+t \\ x^2+9y^2=9 \end{array} \right.\)得\(\left\{\begin{array}{ll}(m^2+9)y^2+2mty+t^2-9=0 \\ (m^2+9)x^2-18tx+9t^2-9m^2=0 \end{array} \right.\)
\(∴y_1+y_2=-\dfrac{2mt}{m^2+9}\)，\(y_1y_2=\dfrac{t^2-9}{m^2+9}\)，\(x_1+x_2=\dfrac{18t}{m^2+9}\)，\(x_1x_2=\dfrac{9t^2-9m^2}{m^2+9}\)
由\(\dfrac{k_{EA_1}}{k_{FA_2}}=\dfrac{k_{PA_1}}{k_{PA_2}}\)得\(\dfrac{\dfrac{y_1}{x_1+3}}{\dfrac{y_2}{x_2-3}}=\dfrac{\dfrac{n}{9}}{\dfrac{n}{3}}=\dfrac{1}{3}\)
\(⇒\dfrac{y_1}{x_1+3}=\dfrac{1}{3} \cdot \dfrac{y_2}{x_2-3} \)
两边平方得\(\dfrac{y_1^2}{(x_1+3)^2}=\dfrac{1}{9} \cdot \dfrac{y_2^2}{(x_2-3)^2}\)
\(⇒\dfrac{1-\dfrac{x_1^2}{9}}{(x_1+3)^2}=\dfrac{1}{9} \cdot \dfrac{1-\dfrac{x_2^2}{9}}{(x_2-3)^2}\)
\(⇒\dfrac{9-x_1^2}{(3+x_1)^2}=\dfrac{1}{9} \cdot \dfrac{9-x_2^2}{(3-x_2)^2}\)
\(⇒\dfrac{3-x_1}{3+x_1}=\dfrac{1}{9} \cdot \dfrac{3+x_2}{3-x_2}\)
\(⇒9(3-x_1)(3-x_2)=(3+x_1)(3+x_2)\)
\(⇒4x_1x_2-15(x_1+x_2)+36=0\)
\(⇒36t^2-36m^2-15×18t+36m^2+36×9=0\)
\(⇒2t^2-15t+18=0\)
\(⇒(2t-3)(t-6)=0\)
显然\(t≠6\)，故\(t=\dfrac{3}{2}\)
故直线\(EF\)恒过定点\((\dfrac{3}{2},0)\)
【解法4】蝴蝶模型斜率比+斜率双用
设\(E(x_1,y_1)\)，\(F(x_2,y_2)\)，\(P(6,n)\)，\(\dfrac{k_{EA_1}}{k|{FA_2}}=\dfrac{k_{PA_1}}{k_{PA_2}}\)
\(⇒\dfrac{\dfrac{y_1}{x_1+3}}{\dfrac{y_2}{x_2-3}}=\dfrac{\dfrac{n}{9}}{\dfrac{n}{3}}=\dfrac{1}{3}\)
\(⇒k_{EA_1}=\dfrac{1}{3} \cdot k_{FA_2}①\)
点差法：\(\left\{\begin{array}{li}\dfrac{x_1^2}{9}+y_1^2=1 \\ \dfrac{x_2^2}{9}+y_2^2=1 \end{array} \right.\)
\(⇒k_{EF}=\dfrac{y_1-y_2}{x_1-x_2}=-\dfrac{1}{9} \cdot \dfrac{x_1+x_2}{y_1+y_2} \)
同理可得：\(k_{EA_1}=\dfrac{y_1}{x_1+3}=-\dfrac{1}{9} \cdot \dfrac{x_1-3}{y_1}② \)
\(k_{FA_2}=\dfrac{y_2}{x_2-3}=-\dfrac{1}{9} \cdot \dfrac{x_2+3}{y_2}③\)
将\(②、③\)代入\(①\)得
\(\left\{\begin{array}{ll}\dfrac{y_1}{x_1+3}=\dfrac{1}{3} \cdot \dfrac{y_2}{x_2-3} \\ -\dfrac{1}{9} \cdot \dfrac{x_1-3}{y_1}=\dfrac{1}{3} \cdot (-\dfrac{1}{9} \cdot \dfrac{x_2+3}{y_2}) \end{array} \right.\)
\(⇒\left\{\begin{array}{li}3y_1x_2-9y_1=y_2x_1+3y_2④ \\ 3y_2x_1-9y_2=y_1x_2+3y_1⑤ \end{array} \right.\)
由\(④-⑤\)得\(4(y_1x_2-y_2x_1)=-6(y_2-y_1)\)，即\(y_1x_2-y_2x_1=-\dfrac{3}{2}(y_2-y_1)⑥\)
由直线\(EF：\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}\)，整理得\(y_1x_2-y_2x_1=y(x_2-x_1)-x(y_2-y_1)⑦\)
将\(⑥\)代入\(⑦\)得直线\(EF：y(x_2-x_1)=(x-\dfrac{3}{2})(y_2-y_1)\)
故直线\(EF\)恒过定点\((\dfrac{3}{2},0)\)
【解法五】蝴蝶模型斜率比+齐次化联立
设\(E(x_1,y_1)\)，\(F(x_2,y_2)\)，\(P(6,n)\)，
由\(\dfrac{k_{EA_1}}{k_{FA_2}}=\dfrac{k_{PA_1}}{k_{PA_2}}\)得\(\dfrac{\dfrac{y_1}{x_1+3}}{\dfrac{y_2}{x_2-3}}=\dfrac{\dfrac{n}{9}}{\dfrac{n}{3}}=\dfrac{1}{3}\)
\(⇒k_{EA_1}=\dfrac{1}{3} \cdot k_{FA_2}①\)
\(k_{EA_1} \cdot k_{EA_2}=\dfrac{y_1}{x_1+3} \cdot \dfrac{y_1}{x_1-3}=\dfrac{y_1^2}{x_1^2-9}=\dfrac{1-\dfrac{x_1^2}{9}}{x_1^2-9}=-\dfrac{1}{9}②\)
由\(①、②\)得\(k_{FA_2} \cdot k_{EA_2}=-\dfrac{1}{3}\)
设\(EF：m(x-3)+ny=1\)
\(x^2+9y^2=9\) \(⇒[(x-3)+3]^2+9y^2=9\)
\(⇒(x-3)^2+6(x-3)[m(x-3)+ny]+9y^2=0\)
\(⇒1+6[m+\dfrac{ny}{x-3}]+9\dfrac{y}{(x-3)^2}=0\)
设\(k=\dfrac{y}{x-2}\)，则\(1+6(m+nk)+9k^2=0\)即\(9k^2+6nk+6m+1=0③\)
设\(k_1=k_{EA_2}=\dfrac{y_1}{x_1-2}\)，\(k_2=k_{FA_2}=\dfrac{y_2}{x_2-2}\)，则\(k_1，k_2\)是方程\(③\)的两根
\(∴k_1k_2=\dfrac{6m+1}{9}=-\dfrac{1}{3}\)
解得\(m=-\dfrac{2}{3}\)
\(∴EF：-\dfrac{2}{3}(x-3)+ny=1\)
令\(y=0\)得\(x=\dfrac{3}{2}\)
故直线EF恒过定点\(\dfrac{3}{2},0)\)
原创文章，作者：leopold，如若转载，请注明出处：https://www.math211.com/2024/02/12/908/