直线l 过点F2(c,0)及点Q（a,b）,
∴l:x=(a-c)y/b+c,①
代入椭圆方程：b^2x^2+a^2y^2=a^2b^2得
(a-c)^2y^2+2bc(a-c)y+b^2c^2+a^2y^2=a^2b^2,
整理得（2a^2-2ac+c^2)y^2+2bc(a-c)y-b^4=0,
设M(x1,y1),N(x2,y2),l上的点R((a-c)m+c,bm),则
y1+y2=-2bc(a-c)/(2a^2-2ac+c^2),y1y2=-b^4/(2a^2-2ac+c^2)
向量MR·NR= -2,得
((a-c)m+c-x1,bm-y1)*((a-c)m+c-x2,bm-y2)
=[(a-c)m+c-x1][(a-c)m+c-x2]+(bm-y1)(bm-y2)
=(a-c)^2(m-y1/b)(m-y2/b)+(bm-y1)(bm-y2)(由①）
=[(a-c)^2/b^2+1][b^2m^2-bm(y1+y2)+y1y2]
=[(a-c)^2/b^2+1][b^2m^2(2a^2-2ac+c^2)+2b^2cm(a-c)-b^4]/(2a^2-2ac+c^2)=-2,
∴m^2(2a^2-2ac+c^2)+2cm(a-c)-b^2=-2(2a^2-2ac+c^2)/[(a-c)^2+b^2],②
OR^2=[(a-c)m+c]^2+b^2m^2
=[(a-c)^2+b^2]m^2+2c(a-c)m+c^2，③
在约束条件②之下求③的最小值，繁！