3344001212
k k k k b b +≥+≤-∴<≤-≤< ································································ 9分 （Ⅲ）假设存在符号条件的点(,0)M m ，则由（Ⅱ）得：
11221212
(,),(,)()()MP x m y MQ x m y MP MQ x m x m y y =-=-?=-?-+ 2121212()x x m x x m y y =-+++ ···················································· 10分 2212121212(1)(1)[()1]y y k x x k x x x x =--=-++2
2943
k k =-+ ·························· 11分 所以222
222241289434343
k k k MP MQ m m k k k -?=-?+-+++ 2222(485)(312)43
m m k m k --+-=+ ············································· 12分 设MP MQ λ?=
即2222(485)(312)43
m m k m k λ--+-=+ 2222(485)(312)43m m k m k λλ--+-=+
对于任意实数(0)k k ≠，上式恒成立，
所以2248543123m m m λλ
?--=??-=?? ·········································································· 13分 得11813564m λ?=????=-??
所以符合条件的点M存在，其坐标为
11
(,0)
8
················································ 14分
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