抛物线y=ax^2+(a+m)x-1/2m过A(1,0)、B（X2,0）,交Y正半轴于C,且S三角形ABC=1/2.

Question

抛物线y=ax^2+(a+m)x-1/2m过A(1,0)、B（X2,0）,交Y正半轴于C,且S三角形ABC=1/2.

Answer 1

抛物线Y过A(1,0)、B（X2,0）,
设,抛物线解析式为Y=a(x-1)(x-x2),有
Y=a[x^2-(1+x2)x+x2].
x1+x2=(1+x2),
x1*x2=x2.
又∵y=ax^2+(a+m)x-m/2,有
x1+x2=-(a+m)/a,
x1*x2=-m/2a,则有
-(a+m)/a=(1+x2),
-m/2a=x2,解方程得,
m=-4a,
x2=-m/2a=2.
抛物线y=ax^2+(a+m)x-1/2m,交Y正半轴于C,则C点坐标为(0,-m/2),且m