Question

已知：经过点P（2,0）斜率为4/3的直线和抛物线：y的平方=2X交于A、B两点,设线段AB中点为M,求：点M坐标.

Answer 1

y = 4(x - 2)/3

y*y = 2x

2x = 16(x - 2)^2/9

8x^2 - 32x + 32 - 9x = 0

8x^2 - 41x + 32 = 0

x1 + x2 = 41/8

则 M 坐标 (X,Y),X = (x1+x2)/2 = 41/16

Y = 4(X-2)/3 = 3/4