题目

直线l:y=kx-2,抛物线c:y^2=4x(k不等于0),求他们相交所得弦中点轨迹方程
具体过程最好说明此类问题的通用解法

提问时间：2021-03-27

答案

设交点A(a²/4,a),B(b²/4,b)
直线方程:(y - b)/(a - b) = (x - b²/4)/(a²/4 - b²/4),y - b = (4x - b²)/(a + b)
直线过C(0,-2):-2 -b = -b²/(a + b)
ab = -2(a + b) (1)
设弦中点M(x,y):
y = (a+ b)/2,a + b = 2y (2)
x = (a²/4 + b²/4)/2,a² + b² = 8x (3)
a² + b² = (a + b)² - 2ab = (2y)² - 2[-2(a + b)] = 4y² + 4(a + b) = 4y² + 8y = 8x
y² + 2y = 2x