( I)解:由題意,|QA|+|QB|=|QP|+|QB|=6,
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∴Q點軌跡是以A、B為焦點的橢圓,且a=3,c=2,
∴曲線C的軌跡方程是
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.(5分)
( II)證明:先考慮切線的斜率存在的情形.設切線l:y=kx+m,則
由l與⊙O相切得
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即m
2=r
2(1+k
2)①(7分)
由
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,消去y得,(5+9k
2)x
2+18kmx+9(m
2-5)=0,
設M(x
1,y
1),N(x
2,y
2),則由韋達定理得
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,
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(9分)
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=
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=
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=
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②(10分)
由于其中一條切線滿足∠MON>90°,對此
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=
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結合①式m
2=r
2(1+k
2)可得
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(12分)
于是,對于任意一條切線l,總有
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,進而
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=
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故總有∠MON>90°.(14分)
最后考慮兩種特殊情況:
(1)當滿足∠MON>90°的那條切線斜率不存在時,切線方程為x=±r.代入橢圓方程可得交點的縱坐標
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,
因∠MON>90°,故
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,得到
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,同上可得:任意一條切線l均滿足∠MON>90°;
(2)當滿足∠MON>90°的那條切線斜率存在時,
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,
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,對于斜率不存在的切線x=±r也有∠MON>90°.
綜上所述,命題成立.(15分)
分析:( I)由題意,|QA|+|QB|=|QP|+|QB|=6,所以Q點軌跡是以A、B為焦點的橢圓,故可求曲線C的軌跡方程;
( II)先考慮切線的斜率存在的情形.設切線l:y=kx+m,利用l與⊙O相切,建立方程,再由
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,消去y,借助于韋達定理,證明
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=
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即可,再考慮兩種特殊情況:(1)當滿足∠MON>90°的那條切線斜率不存在時,切線方程為x=±r,(2)當滿足∠MON>90°的那條切線斜率存在時,故結論可證.
點評:本題考查曲線軌跡的求解,考查橢圓的標準方程,考查直線與圓、橢圓的位置關系,考查學生分析解決問題的能力,需要一定的基本功.