(Ⅰ)设M(x,y),根据条件,可求得x2+3y2=1,从而可证得结论;
(Ⅱ)设|OA|=p,|OB|=q,∠xOA=α,可求得A,B两点的坐标,代入x2+3y2=1,可整理得,应用基本不等式可求得,从而,问题解决.
证明:(Ⅰ)∵,
∴x1x2+3y1y2=0,
又,
设M(x,y),则(x,y)=(x1cosθ,y1cosθ)+(x2sinθ,y2sinθ)=(x1cosθ+x2sinθ,y1cosθ+y2sinθ),
则x2+3y2=(x1cosθ+x2sinθ)2+3(y1cosθ+y2sinθ)2=(x12+3y12)cos2θ+(x22+3y22)sin2θ+2sinθcosθ(x1x2+3y1y2)
∵A(x1,y1),B(x2,y2)是椭圆x2+3y2=1上的两点,
∴x12+3y12=1,x22+3y22=1,又x1x2+3y1y2=0,
∴(x12+3y12)cos2θ+(x22+3y22)sin2θ+2sinθcosθ(x1x2+3y1y2)=cos2θ+sin2θ=1.故点M在椭圆上.
(Ⅱ)设|OA|=p,|OB|=q,∠xOA=α,
则
则
从而,
故.
∴.
的最小值为.