解:(I)令P(x
1,y
1),,Q(x
2,y
2),由題意,可設(shè)拋物線方程為 y
2=2px
由橢圓的方程可得F
1 (-1,0),F(xiàn)
2 (1,0 )故p=2,曲線C的方程為 y
2=4x,
由題意,可設(shè)PQ的方程 x=my-1 (m>0).把PQ的方程代入曲線C的方程 化簡(jiǎn)可得 y
2-4my+4=0,
∴y
1+y
2=4m,y
1y
2=4. 又
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=
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,∴x
1+1=λ(x
2+1),y
1=λy
2,
又
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=λ+
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+2=4m
2.λ∈[2,4],∴2+
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≤λ+
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≤4+
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,
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≤m
2≤
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,
∴
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≤
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≤
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∴直線L的斜率k的取值范圍為[
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,
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].
(II)由于P,M關(guān)于X軸對(duì)稱,故M(x
1,-y
1),,
∵
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-
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=
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+
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=
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=0,
∴M、Q、F
2三點(diǎn)共線,故直線MQ過(guò)定點(diǎn) F
2 (1,0 ).
分析:(I)求出曲線C的方程,把PQ的方程 x=my-1 (m>0)代入曲線C的方程 化簡(jiǎn)可得 y
2-4my+4=0,利用根與系數(shù)的關(guān)系 及
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=
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,可得
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=λ+
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+2=4m
2,據(jù)λ∈[2,4],求得直線L的斜率
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的范圍.
(II)根據(jù)
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-
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=0,可得 M、Q、F
2三點(diǎn)共線,故直線MQ過(guò)定點(diǎn) F
2 (1,0 ).
點(diǎn)評(píng):本題考查橢圓、拋物線的標(biāo)準(zhǔn)方程、簡(jiǎn)單性質(zhì),三點(diǎn)共線的條件,根據(jù)題意,得到2+
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≤λ+
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≤4+
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,是解題的關(guān)鍵.