以极点极线为背景的蝴蝶模型五种解法

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.

以极点极线为背景的蝴蝶模型五种解法

【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/

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