在四邊形ABCD中,已知A(0,0),D(0,4),點(diǎn)B在x軸上,BC∥AD,且對角線AC⊥BD.
(Ⅰ)求點(diǎn)C的軌跡方程;
(Ⅱ)若點(diǎn)P是直線y=2x-5上任意一點(diǎn),過點(diǎn)P作點(diǎn)C的軌跡的兩切線PE、PF,E、F為切點(diǎn),M為EF的中點(diǎn).求證:PM⊥x軸;
(Ⅲ)在(Ⅱ)的條件下,直線EF是否恒過一定點(diǎn)?若是,請求出這個定點(diǎn)的坐標(biāo);若不是,請說明理由.
【答案】
分析:(Ⅰ)設(shè)點(diǎn)C的坐標(biāo)為(x,y),再由共線向量定理求解.(Ⅱ)對函數(shù)
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求導(dǎo)得
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.設(shè)切點(diǎn)坐標(biāo),得切線方程.又設(shè)點(diǎn)P的坐標(biāo)為(t,2t-5),由切線過點(diǎn)P,得E,F(xiàn)所在的直線方程,由韋達(dá)定理求得M坐標(biāo)得證.(Ⅲ)先求得直線AB的方程為:

,即t(x-4)+10-2y=0.(*)當(dāng)x=4,y=5時,方程(*)恒成立,
解答:解:(Ⅰ)如圖,設(shè)點(diǎn)C的坐標(biāo)為(x,y)(x≠0,y≠0),
則
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,
∵
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,
∴x•(-x)+y•4=0,即
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.
∴所求的軌跡T是除去頂點(diǎn)的拋物線(3分)
(Ⅱ)對函數(shù)
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求導(dǎo)得,
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.
設(shè)切點(diǎn)坐標(biāo)為
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,則過該切點(diǎn)的切線的斜率是
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,
該切線方程是
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.
又設(shè)點(diǎn)P的坐標(biāo)為(t,2t-5),
∵切線過點(diǎn)P,
∴有
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,
化簡,得x
2-2tx
+8t-20=0.(6分)
設(shè)A、B兩點(diǎn)的坐標(biāo)分別為
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、
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,
則x
1、x
2為方程x
2-2tx+8t-20=0的兩根,x
1+x
2=2t,?x
1x
2=8t-20.
∴
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因此,當(dāng)t=0時,直線PM與y軸重合,當(dāng)t≠0時,直線PM與y軸平行(9分)
(Ⅲ)∵
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=
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.
∴點(diǎn)M的坐標(biāo)為
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.
又∵
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.
∴直線AB的方程為:
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,即t(x-4)+10-2y=0.(*)
∵當(dāng)x=4,y=5時,方程(*)恒成立,
∴對任意實數(shù)t,直線AB恒過定點(diǎn),定點(diǎn)坐標(biāo)為(4,5).(14分)
點(diǎn)評:本題主要考查向量法求軌跡方程,導(dǎo)數(shù)法求切線方程以及直線過定點(diǎn)問題.