每日一题[858]共线时的长度转化

如图，已知$A,B$两点在椭圆$C:\dfrac{{{x^2}}}{m} + {y^2} = 1$（$m > 1$），直线$AB$上两个不同的点$P,Q$满足$\left| {AP} \right|:\left| {PB} \right| = \left| {AQ} \right|:\left| {QB} \right|$，且$P$点的坐标为$\left( {1, 0} \right)$，求点$Q$的轨迹方程．

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分析与解　设直线$AB$为$\begin{cases} x = 1 + t,\\y = kt,\end{cases}$ 其中$A,B,Q$对应的参数分别为${t_1},{t_2},{t_0}$，则

$$\left| {\dfrac{{{t_1}}}{{{t_2}}}} \right| = \left| {\dfrac{{{t_0} - {t_1}}}{{{t_0} - {t_2}}}} \right|,$$ 易知 $$\dfrac{{{t_1}}}{{{t_2}}} \ne \dfrac{{{t_0} - {t_1}}}{{{t_0} - {t_2}}}.$$
所以 $$\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{{t_0} - {t_1}}}{{{t_2} - {t_0}}},$$ 即 $${t_0} = \dfrac{{2{t_1}{t_2}}}{{{t_1} + {t_2}}},$$ 联立直线与椭圆，有$\dfrac{1}{m}{\left( {t + 1} \right)^2} + {k^2}{t^2} - 1 = 0$，即 $$\left( {{k^2} + \dfrac{1}{m}} \right){t^2} + \dfrac{2}{m}t + \dfrac{1}{m} - 1 = 0,$$
所以 $${t_0} = \dfrac{{2\left( {\frac{1}{m} - 1} \right)}}{{ - \frac{2}{m}}} = m - 1,$$
于是点$Q$的轨迹方程为$x = m$．