解析几何之“定比点差法”

介绍定比点差法之前，先介绍一些解析几何中的基础知识：

一、定比分点

若$\overrightarrow {AM}=\lambda \overrightarrow {MB}$，则称点$M$为点$A$、$B$的$\lambda $定比分点．

当$\lambda >0$时，点$M$在线段$AB$上，称为内分点；

当$\lambda <0$（$\lambda \ne -1$）时，点$M$在线段$AB$的延长线上，称为外分点．

定比分点坐标公式：若点$A(x_1,y_1)$，$B(x_2,y_2)$，$\overrightarrow {AM}=\lambda \overrightarrow {MB}$，则点$M$的坐标为\[M\left(\dfrac {x_1+\lambda x_2}{1+\lambda },\dfrac {y_1+\lambda y_2}{1+\lambda }\right ).\]

二、点差法

若点$A(x_1,y_1),B(x_2,y_2)$在有心二次曲线$\dfrac {x^2}{a^2}\pm\dfrac {y^2}{b^2}=1$上，则有\[\dfrac {x_1^2}{a^2}\pm\dfrac {y_1^2}{b^2}=1,\dfrac {x_2^2}{a^2}\pm\dfrac {y_2^2}{b^2}=1,\]两式作差得$$\dfrac {(x_1+x_2)(x_1-x_2)}{a^2}\pm \dfrac {(y_1+y_2)(y_1-y_2)}{b^2}=0.$$此即有心二次曲线的垂径定理，可以解决与弦的中点相关的问题．

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下面介绍定比点差法：

若点$A(x_1,y_1),B(x_2,y_2)$在有心二次曲线$\dfrac {x^2}{a^2}\pm\dfrac {y^2}{b^2}=1$上，则有\[\begin{split} \dfrac {x_1^2}{a^2}\pm\dfrac {y_1^2}{b^2}=1,\dfrac {\lambda ^2x_2^2}{a^2}\pm\dfrac {\lambda ^2y_2^2}{b^2}=\lambda ^2,\end{split} \]两式作差得$$\dfrac {(x_1+\lambda x_2)(x_1-\lambda x_2)}{a^2}\pm \dfrac {(y_1+\lambda y_2)(y_1-\lambda y_2)}{b^2}=1-\lambda ^2.$$这样就得到了\[\dfrac {1}{a^2}\cdot\dfrac {x_1+\lambda x_2}{1+\lambda }\cdot\dfrac {x_1-\lambda x_2}{1-\lambda }\pm\dfrac {1}{b^2}\cdot\dfrac {y_1+\lambda y_2}{1+\lambda }\cdot\dfrac {y_1-\lambda y_2}{1-\lambda }=1.\]

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例1　过异于原点的点$P(x_0,y_0)$引椭圆$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1(a>b>0)$的割线$PAB$，其中点$A,B$在椭圆上，点$M$是割线$PAB$上异于$P$的一点，且满足$\dfrac {AM}{MB}=\dfrac {AP}{PB}$．求证：点$M$在直线$\dfrac {x_0x}{a^2}+\dfrac {y_0y}{b^2}=1$上．

证明　直接运用定比点差法即可．

设$\overrightarrow {AP}=\lambda \overrightarrow {PB}$，则有$\overrightarrow {AM}=-\lambda \overrightarrow {MB}$，设$A(x_1,y_1),B(x_2,y_2),M(x_M,y_M)$，则有\[\begin{split} x_0=\dfrac {x_1+\lambda x_2}{1+\lambda },y_0=\dfrac {y_1+\lambda y_2}{1+\lambda }.\\x_M=\dfrac {x_1-\lambda x_2}{1-\lambda },y_M=\dfrac {y_1-\lambda y_2}{1-\lambda }.\end{split} \]又因为点$A,B$在椭圆上，所以有\[\begin{split} \dfrac {x_1^2}{a^2}+\dfrac {y_1^2}{b^2}=1,\dfrac {\lambda ^2x_2^2}{a^2}+\dfrac {\lambda ^2y_2^2}{b^2}=\lambda ^2,\end{split} \]两式作差得\[\dfrac {(x_1+\lambda x_2)(x_1-\lambda x_2)}{a^2}+\dfrac {(y_1+\lambda y_2)(y_1-\lambda y_2)}{b^2}=1-\lambda ^2.\]两边同除以$1-\lambda ^2$，即可得到\[\dfrac {x_0x_M}{a^2}+\dfrac {y_0y_M}{b^2}=1.\]命题得证．

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练习1　（2008高考数学安徽卷理科）设椭圆$C$：$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1(a>b>0)$过点$M(\sqrt 2,1)$，且焦点为$F_1(-\sqrt 2,0)$．

（1）求椭圆的方程；

（2）过点$P(4,1)$的动直线$l$与椭圆$C$相交于不同点$A,B$时，在线段$AB$上取点$Q$，满足$|AP|\cdot|QB|=|AQ|\cdot|PB|$，证明：点$Q$总在某定直线上．

答案　（1）$\dfrac {x^2}{4}+\dfrac {y^2}{2}=1$；（2）点$Q$在直线$2x+y-2=0$上．

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例2　已知椭圆$\dfrac {x^2}{9}+\dfrac {y^2}{4}=1$，过定点$P(0,3)$的直线与椭圆交于两点$A,B$（$A,B$可以重合），求$\dfrac {PA}{PB}$的取值范围．

解　设$A(x_1,y_1),B(x_2,y_2)$，$\overrightarrow {AP}=\lambda \overrightarrow {PB}$，则$\dfrac {PA}{PB}=-\lambda$．

于是$P\left(\dfrac {x_1+\lambda x_2}{1+\lambda },\dfrac {y_1+\lambda y_2}{1+\lambda }\right ) =(0,3)$，于是$$\begin{eqnarray} x_1+\lambda x_2=0,y_1+\lambda y_2=3(1+\lambda ).\end{eqnarray} $$又因为点$A,B$在椭圆上，所以有\[\begin{split} \dfrac {x_1^2}{9}+\dfrac {y_1^2}{4}=1,\dfrac {\lambda ^2x_2^2}{9}+\dfrac {\lambda ^2y_2^2}{4}=\lambda ^2,\end{split} \]两式相减得\[\begin{eqnarray} \dfrac {(x_1+\lambda x_2)(x_1-\lambda x_2)}{9}+ \dfrac {(y_1+\lambda y_2)(y_1-\lambda y_2)}{4}=1-\lambda ^2.\end{eqnarray} \]将(1)代入(2)中得到\[\begin{eqnarray} y_1-\lambda y_2=\dfrac 43(1-\lambda ).\end{eqnarray} \]由(1)(3)解得\[y_1=\dfrac {3(1+\lambda )+\frac 43(1-\lambda )}{2}=\dfrac {13}{6}+\dfrac 56\lambda \in [-2,2].\]从而解得$\lambda $的取值范围为$\left[-5,-\dfrac 15\right ]$，于是$\dfrac {PA}{PB}$的取值范围为$\left[\dfrac 15,5\right ]$． 

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练习2　设$D(0,16)$，$M,N$是椭圆$\dfrac {x^2}{25}+\dfrac {y^2}{16}=1$上的两个动点（可以重合），且$\overrightarrow {DM}=\lambda \overrightarrow {DN}$，求实数$\lambda $的取值范围．

答案　$\left[\dfrac 35,\dfrac 53\right ]$．

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例3　设$F_1(-c,0)$、$F_2(c,0)$为椭圆$\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}=1(a>b>0)$的左、右焦点，$P$为椭圆上任意一点，直线$PF_1,PF_2$分别交椭圆于异于$P$的点$A$、$B$，若$\overrightarrow {PF_1}=\lambda \overrightarrow {F_1A}$，$\overrightarrow {PF_2}=\mu\overrightarrow {F_2B}$，求证：$\lambda +\mu=2\cdot\dfrac {a^2+c^2}{a^2-c^2}$．

证明　设$P(x_0,y_0)$，$A(x_1,y_1)$，$B(x_2,y_2)$，则$$F_1\left(\dfrac {x_0+\lambda x_1}{1+\lambda },\dfrac {y_0+\lambda y_1}{1+\lambda }\right ),F_2\left(\dfrac {x_0+\mu x_2}{1+\mu},\dfrac {y_0+\mu y_2}{1+\mu}\right ).$$于是有\[\begin{split} \begin{eqnarray} x_0+\lambda x_1=-(1+\lambda )c,y_0+\lambda y_1=0;\\x_0+\mu x_2=(1+\mu )c,y_0+\mu y_2=0.\end{eqnarray} \end{split} \]又由点$P,A$在椭圆上得到\[\begin{split} \dfrac {x_0^2}{a^2}+\dfrac {y_0^2}{b^2}=1,\dfrac {\lambda ^2x_1^2}{a^2}+\dfrac {\lambda ^2y_1^2}{b^2}=\lambda ^2,\end{split} \]两式相减得\[\begin{eqnarray} \dfrac {(x_0+\lambda x_1)(x_0-\lambda x_1)}{a^2}+ \dfrac {(y_0+\lambda y_1)(y_0-\lambda y_1)}{b^2}=1-\lambda ^2.\end{eqnarray} \]从而有\[x_0-\lambda x_1=\dfrac {a^2}{c}(\lambda -1).\]结合(4)式可解得\[2x_0=\dfrac {a^2}{c}(\lambda -1)-c(1+\lambda ).\]同理可得\[x_0-\mu x_2=\dfrac {a^2}{c}(1-\mu ).\]结合(5)式得到\[2x_0=\dfrac {a^2}{c}(1-\mu)+c(1+\mu).\]于是有\[\dfrac {a^2}{c}(\lambda -1)-c(1+\lambda )=\dfrac {a^2}{c}(1-\mu)+c(1+\mu).\]整理得$\lambda +\mu=2\cdot\dfrac {a^2+c^2}{a^2-c^2}$，命题得证．

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练习3　已知过椭圆$\dfrac {x^2}{2}+y^2=1$的左焦点$F$的直线交椭圆于$A,B$两点，且有$\overrightarrow {FA}=3\overrightarrow {BF}$，求点$A$的坐标．

答案　$A(0,\pm 1)$．

定比点差法实际上是直线的参数方程的变异形式，只不过将其中的$t$变作了$\lambda $，也就是说只要是共线点列的问题都可以在考虑运用直线的参数方程的同时考虑定比点差法．定比点差法在处理圆锥曲线上过定点的直线的证明题时往往可以起到简化运算的作用．但定比点差法无法应用于抛物线，并且它采用的参数$\lambda $在解析几何问题中并不通用，在求解具体的斜率、弦长与面积时往往会引起运算上的麻烦（当然，求坐标还是很简便的），所以并不是所有的共线问题都适用用定比点差法解决．