25~26高二上·四川成都七中期末·第14题

$x – y – 2 = 0$

【题目】已知抛物线$x^2 = 4y$，过动点$P$作抛物线的两条切线，记两条切线的斜率分别为$k_1$、$k_2$，若$(k_1 – 1)(k_2 – 1) + 1 = 0$，则动点$P$的轨迹方程为______.

【答案】$x – y – 2 = 0$

【解析】设$P(x_0, y_0)$，切线方程为$y = k(x – x_0) + y_0$，

$y = k(x – x_0) + y_0$与$x^2 = 4y$联立得$x^2 – 4kx + 4kx_0 – 4y_0 = 0$，
由$\Delta = 0$，整理得$k^2 – x_0k + y_0 = 0$，
则方程$k^2 – x_0k + y_0 = 0$的两个根为$k_1$和$k_2$，
则$k_1 + k_2 = x_0$，$k_1k_2 = y_0$.

$\because (k_1 – 1)(k_2 – 1) + 1 = 0$，
$\therefore k_1k_2 – (k_1 + k_2) + 1 + 1 = 0$，
$\therefore y_0 – x_0 + 2 = 0$，
$\therefore x_0 – y_0 – 2 = 0$，

$\therefore$ 动点$P$的轨迹方程为$x – y – 2 = 0$.

故答案为：$x – y – 2 = 0$