解:(Ⅰ)設(shè)點(diǎn)C(x,y),由題意得H(x,
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y),
則
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,由于AC⊥BH,
于是
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,
又y=0時
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共線,不合題意.故點(diǎn)C的軌跡方程為
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(y≠0).
設(shè)點(diǎn)H(x,y),C(x
0,y
0),則
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(y
0≠0),
由
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得到點(diǎn)H的軌跡方程為
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.(4分)
(Ⅱ)設(shè)
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,則
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,

,
故
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=
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,
所以
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不能構(gòu)成等差數(shù)列.(9分)
(Ⅲ)設(shè)M(9,m),N(9,n),則A(-3,0),B(3,0),
于是
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由A,H,M三點(diǎn)共線得
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,∴
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;
由B,H,N三點(diǎn)共線得
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,又
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,以MN為直徑的圓的方程為
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解得
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(舍)或
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.故以MN為直徑的圓必過橢圓外定點(diǎn)(17,0).(15分)
分析:(Ⅰ)設(shè)點(diǎn)C(x,y),由題意得H(x,

y),則

,由于AC⊥BH,于是

,又y=0時

共線,不合題意.故點(diǎn)C的軌跡方程為

(y≠0).由此能得到得到點(diǎn)H的軌跡方程為

.
(Ⅱ)設(shè)
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,則
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,
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,由此能得到
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不能構(gòu)成等差數(shù)列.
(Ⅲ)設(shè)M(9,m),N(9,n),則A(-3,0),B(3,0),于是

,由A,H,M三點(diǎn)共線得

.由B,H,N三點(diǎn)共線得
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,又
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,以MN為直徑的圓的方程為

,由此能得以MN為直徑的圓必過橢圓外定點(diǎn)(17,0).
點(diǎn)評:本題考查直線與圓錐曲線的綜合運(yùn)用,解題時要認(rèn)真審題,仔細(xì)解答,注意挖掘題設(shè)中的隱含條件.