每日一题[643]定比点差法

过点$P(3,1)$的动直线$l$与双曲线$C:\dfrac{x^2}3-y^2=1$的左、右两支分别交于点$A,B$，在线段$AB$上取不同于$A,B$的点$Q$，满足$|AP|\cdot |QB|=|AQ|\cdot |PB|$，求证：点$Q$总在某条定直线上．

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分析与解　由于$|AP|\cdot |QB|=|AQ|\cdot |PB|$，不妨设$\overrightarrow {AP}=\lambda \overrightarrow {PB}$，$\overrightarrow {AQ}=-\lambda \overrightarrow {QB}$，$A(x_1,y_1)$，$B(x_2,y_2)$，则利用定比点差法，有

$$P\left(\dfrac{x_1+\lambda x_2}{1+\lambda},\dfrac{y_1+\lambda y_2}{1+\lambda}\right),Q\left(\dfrac{x_1-\lambda x_2}{1-\lambda},\dfrac{y_1-\lambda y_2}{1-\lambda}\right),$$ 于是由 $$\dfrac{x_1^2}3-y_1^2=1,\dfrac{\lambda^2x_2^2}{3}-\lambda^2y_2^2=\lambda^2,$$ 两式相减得 $$\dfrac{(x_1+\lambda x_2)(x_1-\lambda x_2)}3-(y_1+\lambda y_2)(y_1-\lambda y_2)=1-\lambda^2,$$ 整理，即 $$\dfrac 13\cdot \dfrac{x_1+\lambda x_2}{1+\lambda}\cdot\dfrac{x_1-\lambda x_2}{1-\lambda}-\dfrac{y_1+\lambda y_2}{1+\lambda}\cdot \dfrac{y_1-\lambda y_2}{1-\lambda }=1,$$
从而 $$\dfrac{x_1-\lambda x_2}{1-\lambda}-\dfrac{y_1-\lambda y_2}{1-\lambda}-1=0,$$ 因此点$Q$在定直线$x-y-1=0$上．