(18)本小题主要考查直线与椭圆的基本知识,考查分析问题和解决问题的能力。满分15分。(Ⅰ)椭圆方程为x2/a2+(y-r)2/b2=1焦点坐标为(Ⅱ)证明:将直线CD的方程y=k?x代入椭圆方程,得b2x2+a2(k1x-r)2=a2b2,整理,得(b2+a2k12)x2-2k1a2rx+(a2r2-a2b2)=0根据韦达定理,得x1+x2=2k1a2r/(b2+a2k12),x1·x2=(a2r2-a2b2)/(b2+a2k12),所以x1x2/(x1+x2)=(r2-b2)/2k1r①将直线GH的方程y=k2x代入椭圆方程,同理可得x3x4/(x3+x4)=(r2-b2)/2k2r②由①,②得k1x1x2/(x1+x2)=(r2-b2/2r=k2x3x4/(x3+x4)所以结论成立。(Ⅲ)证明:设点P(p,o),点Q(q,o)。由C,P,H共线,得(x1-p)/(x4-p)=k1x1/k2x4解得P=(k1-k2)x1x4/(k1x1-k2x4)由D,Q,G共线,同理可得q=(k1-k2)x2x3/(k1x2-k2x3)由k1x1x2/(x1+x2)=k2x3x4/(x3+x4),变形得:x2x3/(k1x2-k2x3)=x1x4/(k1x1-k2x4)即:(k1-k2)x2x3/(k1x2-k2x3)=(k1-k2)x1x4/(k1x1-k2x4)所以|p|=|q|,即,|OP|=|OQ|。