斜率双用处理圆锥曲线中直线过定点问题

【例1】\(M(2,1)、A、B\)在椭圆\(\dfrac{x^{2}}{6}+\dfrac{y^{2}}{3}=1\)上，且\(k_{MA}+k_{MB}=2\)，求证：直线\(AB\)过定点．

【解析】设\(A(x_{1},y_{1})，B(x_{2},y_{2})\)，则\( \left\{ \begin{array}{ll} \dfrac{x_{1}^{2}}{6}+\dfrac{y_{1}^{2}}{3}=1 \\ \dfrac{x_{2}^{2}}{6}+\dfrac{y_{2}^{2}}{3}=1 \end{array} \right.\)

两式相减整理得\(k_{AB}=\dfrac{y_1-y_2}{x_1-x_2}=-\dfrac{1}{2} \cdot \dfrac{x_1+x_2}{y_1+y_2}\)

同理可得：\( \left\{ \begin{array}{ll} k_{MA}=\dfrac{y_1-1}{x_1-2}=-\dfrac{1}{2} \cdot \dfrac{x_1+2}{y_1+1} ①\\ k_{MB}=\dfrac{y_2-1}{x_2-2}=-\dfrac{1}{2} \cdot \dfrac{x_2+2}{y_2+1} ②\end{array} \right.\)

将\(①、②\)代入\(k_{MA}+k_{MB}=2\)得

\( \left\{ \begin{array}{ll} \dfrac{y_1-1}{x_1-2}+(-\dfrac{1}{2} \cdot \dfrac{x_2+2}{y_2+1})=2 \\ \dfrac{y_2-1}{x_2-2}+(-\dfrac{1}{2} \cdot \dfrac{x_1+2}{y_1+1})=2 \end{array} \right.\)

整理得\( \left\{ \begin{array}{ll} 2y_1y_2+2(y_1-y_2)-2-x_1x_2-2(x_1-x_2)+4=4y_2x_1+4x_1-8y_2-8③ \\ 2y_1y_2+2(y_2-y_1)-2-x_1x_2-2(x_2-x_1)+4=4y_1x_2+4x_2-8y_1-8④\end{array} \right.\)

由\(④-③\)得：\(y_1x_2-y_2x_1=-(y_2-y_1)-2(x_2-x_1)⑤ \)

直线MN：\( \dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}\)，整理得\(y_1x_2-y_2x_1=y(x_2-x_1)-x(y_2-y_1)⑥\)

将\(⑤\)代入\(⑥\)整理得\((y+1)(x_2-x_1)=(x-2)(x_2-x_1)\)

令\(x=2\)得\(y=-1\)

故直线\(MN\)恒过定点\((2,-1)\)

【例2】抛物线C:y^(2)=4x，点A(4,4)在抛物线上，M，N为抛物线上两点，若kMA∙kNA=2，求证：直线MN恒过定点．

【解析】（1）设\(M(x_1,y_1)，N(x_2,y_2)，\)则\(\left\{ \begin{array}{ll} y_1^2=4x_1 \\ y_2^2=4x_2 \end{array} \right.\)，

相减整理得\(k_{MN}=\dfrac{y_1-y_2}{x_1-x_2}=\dfrac{4}{y_1+y_2}\)

同理可得：\(\left\{ \begin{array}{ll} k_{MA}=\dfrac{y_1-4}{x_1-4}=\dfrac{4}{y-1+4}① \\ k_{NA}=\dfrac{y_2-4}{x_2-4}=\dfrac{4}{y_2+4}② \end{array} \right.\)

将\(①、②\)代入\(k_{MA} \cdot k_{NA}=2\)得

\(\left\{ \begin{array}{ll} \dfrac{y_1-4}{x_1-4} \cdot \dfrac{4}{y_2+4}=2 \\ \dfrac{y_2-4}{x_2-4} \cdot \dfrac{4}{y_1+4}=2 \end{array} \right.\)

整理得\(\left\{ \begin{array}{ll} y_2x_1+4x_1-2y_1-4y_2-8=0③ \\ y_1x_2+4x_2-2y_2-4y_1-8=0④ \end{array} \right.\)

由\(④-③\)得\(y_1x_2-y_2x_1=-4(x_2-x_1)-2(y_2-y_1)⑤\)

直线\(MN：\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}，\)整理得\(y_1x_2-y_2x_1=y(x_2-x_1)-x(y_2-y_1)⑥\)

将\(⑤\)代入\(⑥\)整理得\((y+4)(x_2-x_1)=(x-2)(x_2-x_1)\)

故直线\(MN\)恒过定点\((2,-4)\)

【例3】（重庆南开中学12月月考）椭圆\(C：\dfrac{x^2}{6}+\dfrac{y^2}{2}=1，F_1、F_2\)分别为左、右焦点，过\(F_1\)作直线\(l_1、l_2\)分别交椭圆与\(A、B\)和\(C、D\)，直线\(l_1，l_2\)的斜率分别为\(k_1，k_2\)，且\(k_1k_2=\dfrac{1}{2}\)，记\(△OMN\)的面积为\(S_1\)，\(△F_2MN\)的面积为\(S_2\)，求证：\(\dfrac{S_1}{S_2}\)为定值，并求出这个定值．

【解析】设\(M(x_1,y_1)，N(x_2,y_2)\)

由点差法可得\(k_{AB}=\dfrac{y_A-y_B}{x_A-x_B}=-\dfrac{1}{3} \cdot \dfrac{x_A+x_B}{y_A+y_B}=-\dfrac{1}{3} \cdot \dfrac{x_1}{y_1}\)

又\(k_{MF_1}=\dfrac{y_1}{x_1+2}=k_{AB}\)，故\(k_1=\dfrac{y_1}{x_1+2}=-\dfrac{1}{3} \cdot \dfrac{x_1}{y_1}①\)

同理可得：\(k_2=\dfrac{y_2}{x_2+2}=-\dfrac{1}{3} \cdot \dfrac{x_2}{y_2}②\)

将\(①、②\)代入\(k_1k_2=\dfrac{1}{2}\)得

\(\left\{ \begin{array}{ll} \dfrac{y_1}{x_1+2} \cdot (-\dfrac{1}{3} \cdot \dfrac{x_2}{y_2})=\dfrac{1}{2} \\ \dfrac{y_2}{x_2+2} \cdot (-\dfrac{1}{3} \cdot \dfrac{x_1}{y_1})=\dfrac{1}{2} \end{array} \right.\)

整理得\(\left\{ \begin{array}{ll} 2y_1x_2+3y_2x_1=-6y_2③ \\ 2y_2x_1+3y_1x_2=-6y_1④ \end{array} \right.\)

由\(④-③\)得：\(y_1x_2-y_2x_1=6(y_2-y_1)⑤\)

直线\(MN：\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}，\)整理得\(y_1x_2-y_2x_1=y(x_2-x_1)-x(y_2-y_1)⑥\)

将\(⑤\)代入\(⑥\)得\(y(x_2-x_1)=(x+6)(y_2-y_1)\)

故直线\(MN\)恒过定点\(P(-6,0)\)

故\(S_1=|S_{△OPM}-S_{△OPN}|=\dfrac{1}{2} \cdot |OP| \cdot |y_2-y_1|=3|y_2-y_1|\)

\(S_2=|S_{△F_2PM}-S_{△F_2PN}|=\dfrac{1}{2} \cdot |PF_2| \cdot |y_2-y_1|=4|y_2-y_1|\)

∴\(\dfrac{S_1}{S_2}=\dfrac{3|y_2-y_1|}{4|y_2-y_1|}=\dfrac{3}{4}\)