1. y=-x+1
2. √2
3. 20
4. (√35/2,√3)
5. (1)设动点P的坐标为(x,y),则点H坐标为(0,y)
向量PM·向量PN=2*向量PH·向量PH,
(-2-x)(2-x)+(0-y)(0-y)=2[(0-x)(0-x)+(y-y)(y-y)]
整理得到点P轨迹C的方程：y²-x²=4
(2)设直线l方程为y=k(x-2),与曲线C方程y²-x²=4联立消去y得：
x²-[4k²/(k²-1)]x+4=0,由韦达定理可知：x1+x2=4k²/(k²-1),x1x2=4
所以y1+y2=k(x1-2)+k(x2-2)=k(x1+x2-4)=4k/(k²-1)
y1y2=k²(x1-2)(x2-2)=k²x1x2-2k²(x1+x2)-4k²=-8k^4/(k²-1)
若要使AB两点都在x轴下方,需要有y1+y20,△=[4k²/(k²-1)]²-16＞0
解不等式得：√2/2