作业帮在直角坐标系中,抛物线y=x^2+mx-3/4m^2(m>0)与x轴交于A,B两点.若A,B两点到原点的距离分别为OA,OB.且满足1/OB-1/OA=2/3,求m.
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 18:19:59
在直角坐标系中,抛物线y=x^2+mx-3/4m^2(m>0)与x轴交于A,B两点.若A,B两点到原点的距离分别为OA,OB.且满足1/OB-1/OA=2/3,求m.
在直角坐标系中,抛物线y=x^2+mx-3/4m^2(m>0)与x轴交于A,B两点.若A,B两点到原点的距离分别为OA,OB.且满足1/OB-1/OA=2/3,求m.
在直角坐标系中,抛物线y=x^2+mx-3/4m^2(m>0)与x轴交于A,B两点.若A,B两点到原点的距离分别为OA,OB.且满足1/OB-1/OA=2/3,求m.
y=x^2+mx-3/4m^2
=x^2+mx+(m/2)^2-(m/2)^2-3/4m^2
=(x+m/2)^2-1/4m^2-3/4m^2
=(x+m/2)^2-m^2
=(x+m/2+m)(x+m/2-m)
=(x+3m/2)(x-m/2)
令y=0,则(x+3m/2)(x-m/2)=0
解得x1=—3m/2,x2=m/2
所以 OA=|x1|=3m/2
OB=|x2|=m/2
当OA=3m/2,OB=m/2时
1/OB-1/OA=2/3
（代入过程略）
解得m=2
抛物线方程化为y=(x+3/2*m)*(x-1/2*m)
OA、OB分别为1/2*m、3/2*m带入等式得m=4
(1）y=x^2+mx-3/4m^2
=x^2+mx+1/4m^2-m^2
=(x+1/2m)^2-m^2
抛物线对称轴为x=-1/2m
由m>0知-1/2m<0,即对称轴在y轴左侧
(2)y=x^2+mx-3/4m^2
=(x+3/2m)(x-1/2m)
全部展开
(1）y=x^2+mx-3/4m^2
=x^2+mx+1/4m^2-m^2
=(x+1/2m)^2-m^2
抛物线对称轴为x=-1/2m
由m>0知-1/2m<0,即对称轴在y轴左侧
(2)y=x^2+mx-3/4m^2
=(x+3/2m)(x-1/2m)
由1/OB-1/OA=2/3>0可知，Xa
带入1/OB-1/OA=2/3得 2/m+2/3m=2/3
解得m=4,则抛物线解析式为y=x^2+4x-3
收起