（1）焦点坐标 F1(-c,0),F2(c,0),
离心率 e = √2/2 = c/a,a = √2c
右准线为 x = a²/c = 2c
点M,N坐标 M(2c,y1),N(2c,y2).
F1M = (3c,y1)
F2N = (c,y2)
由 F1M·F2M=0,|F1M|=|F2N|=2√5
得 3c²＋y1y2=0,√9c²＋y1²=√c²＋y2²=2√5
解得 c=√2
所以 a = √2c = 2,b = √a² - c² = √2
（2）MN = (0,y2 - y1),
|MN| = |y2 - y1|= |y1|＋|y2| ≥ 2√|y1|·|y2|
当|y1| = |y2|时,|MN|取最小值,此时 y1 = - y2
F1M＋F2N = (4c,y1＋y2) = (4c,0)
F1F2 = (2c,0)
F1M＋F2N = 2·F1F2
所以共线.