解:(Ⅰ)所求椭圆M的方程为…3分 (Ⅱ)当≠,设直线AB的斜率为k = tan,焦点F ( 3 , 0 ),则直线AB的方程为 y = k ( x – 3 ) 有( 1 + 2k2 )x2 – 12k2x + 18( k2 – 1 ) =" 0" 设点A ( x1 , y1 ) , B ( x2 , y2 ) 有x1 + x2 =, x1x2 = |AB| = 又因为k = tan=代入**式得 |AB| = 当=时,直线AB的方程为x = 3,此时|AB| = 而当=时,|AB| == |AB| = 同理可得 |CD| == 有|AB| + |CD| =+= 因为sin2∈[0,1],所以 当且仅当sin2=1时,|AB|+|CD|有最小值是
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