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

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

弦的中点问题通常都用点差法

直线l:y=kx-2,抛物线c:y^2=4x(k不等于0)，则（x1+x2）/2=(4k-1)/4k², (y1+y2)/2=k/4 ,
弦的中点（x1+x2）/2，(y1+y2)/2）＝（(4k-1)/4k²,k/4 ）在直线y=kx-2上，
所以k/4＝k（(4k-1)/4k²)-2
∴