Question

已知抛物线y=(k-1)x²+2kx+k+1,若抛物线与x轴交于A、B,与Y轴交于点C,且△ABC的面积为4,

Answer 1

1) 在抛物线y=(k-1)x²+2kx+k+1中,令 x = 0,即可获得函数与 y 轴的

交点 C 的坐标（0,k+1）,也就是三角形ABC的高为|k+1|

2) 用求根公式求出y=(k-1)x²+2kx+k+1 的两个根,即与 x 轴的两个交点坐标

△ = (2k)^2 - 4(k-1)(k+1)

= 1

x1 = (-2k-1)/[2(k-1)]

x2 = (-2k+1)/[2(k-1)]

x 轴上两交点的距离AB = |x1-x2| = |1/(1-k)|

3)因为△ABC的面积为 4 ,则

|k+1|*|1/(1-k)|/2 = 4

（k+1）/(1-k)/2 = 4 或 （k+1）/(1-k)/2 = -4

k1 = 7/9 k2 = 9/7