已知F1（-2,0）,F2（2,0）,点P满足|PF1|-|PF2|=2,设P的轨迹为E,若直线L过点F2,且法向量为（a,1)(a为实

c=2,a=1,b=√3
点P轨迹为双曲线,方程为 x^2-y^2/3=1
设直线L方程为y=k(x-2)
直线的法向量为(a,1),即其垂线斜率为k1=1/a
∴直线斜率为k=-1/k1=-a
∴直线方程为y=-a(x-2)
设P(x1,y1),Q(x2,y2),M(m,0),则
MP与MQ垂直恒成立,则有
k(MP)*k(MQ)=-1,即y1/(x1-m)*y2/(x2-m)=-1
即y1y2=-(x1-m)(x2-m)=-x1x2+m(x1+x2)-m^2 (1)
将y=-a(x-2)代入双曲线得
x^2-a^2(x-2)^2/3=1,整理得
(3-a^2)x^2+4a^2x-(4a^2+3)=0
有x1+x2=4a^2/(a^2-3),x1x2=(4a^2+3)/(a^2-3)
∴y1y2=a^2(x1-2)(x2-2)=a^2x1x2-2a^2(x1+x2)+4a^2 (2)
(1)=(2),有
(a^2+1)x1x2-(2a^2+m)(x1+x2)+4a^2+m^2=0
代入x1,x2,得
(a^2+1)*(4a^2+3)/(a^2-3)-(2a^2+m)*4a^2/(a^2-3)+4a^2+m^2=0
(a^2+1)*(4a^2+3)-(2a^2+m)*4a^2+(4a^2+m^2)*(a^2-3)=0
4a^4+7a^2+3-8a^4-4ma^2+4a^4+m^2a^2-12a^2-3m^2=0
(a^2-3)m^2-4a^2m-(5a^2-3)=0
△=16a^4+4(a^2-3)(5a^2-3)
=36a^4-4*18a^2+4*9
=36(a^2-1)^2
≥0
即上述方程的解存在,即存在定点M使MP与MQ垂直恒成立
解上述二次方程可得m=(2a^2±3|a^2-1|)/(a^2-3)
无论a取何值,均可解得m=-1或(5a^2-3)/(a^2-3)
(特殊地,当a^2=1时,m的两个解重合)
∴ 点M坐标为M(-1,0)或M((5a^2-3)/(a^2-3),0)