1.已知抛物线y=x^2+mx-2m^2(m≠0)

问题描述:

1.已知抛物线y=x^2+mx-2m^2(m≠0)
(1)求证:该抛物线与x轴有两个不同交点.
(2)过点P(0,n)作y轴的垂线交该抛物线与点A和点B(点A在点P的左边),是否存在实数m,n,使得AP=2PB?若存在,则求出m,n满足的条件;若不存在,请说明理由.
2.已知:二次函数y=x^2-(m+1)x+m的图像交x轴于A(x1,0),B (x2,0),交y轴的正半轴于点C,且x1^2+x2^2=10,求此二次函数的解析式.

(1)x^2+mx-2m^2 = 0
delt = m^2 + 8m^ = 10m^2 >0
(2)m0
y=n时,x^2+mx-2m^2 = n, (x+m/2)^2 = n+(9m^2)/4 > 0,即n>-(9m^2)/4
假设x1m/2时,n + 2m^2>0, n>-2m^2
2PB = 2|x2| =-m + √[n+(9m^2)] = AP = m/2 + √[n+(9m^2)/4]
3m = √[n+(9m^2)], n=0,满足不等式条件,此时要求m>0,两交点分别在y轴两侧
当 √[n+(9m^2)/4]