|
即△=4p2-8p=4p(p-2)=0,得:p=2为所求;
(2)抛物线y2=4x的准线x=-1.
且|AF|+|BF|=8,由定义得x1+x2+2=8,则x1+x2=6
设C(m,0),由C在AB的垂直平分线上,从而|AC|=|BC|
则(x1−m)2+y12=(x2−m)2+y22
(x1−m)2−(x2−m)2=−y12+y22
(x1+x2-2m)(x1-x2)=-4(x1-x2)
因为x1≠x2,所以x1+x2-2m=-4
又因为x1+x2=6,所以m=5,则点C的坐标为(5,0);
(3)设AB的中点M(x0,y0),有x0=
x1+x2 |
2 |
设直线l方程y=k(x-5)过点M(3,y0),得y0=-2k
又因为点M(3,y0)在抛物线y2=4x的内部,则y02<12
得:4k2<12,则k2<3
又因为x1≠x2,则k≠0
故k的取值范围为(−
3, |
3 |