解:(1)易知拋物線的頂點(diǎn)D(-6,-3),則DE=3,OE=6;
∵AE
2=3DE=9,
∴AE=3,即A(-3,0);
將A點(diǎn)坐標(biāo)代入拋物線的解析式中,
得:a(-3+6)
2-3=0,
即a=
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,
即拋物線的解析式為:y=
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(x+6)
2-3=
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x
2+4x+9;
(2)如圖2,連接AP.設(shè)P(-6,a),
∵A(-3,0),E(-6,0),C(0,9),
根據(jù)勾股定理,AE
2+PE
2+AC
2=PC
2,
即9+a
2+9+81=6
2+(a-9)
2,
解得a=1,
P點(diǎn)坐標(biāo)為(-6,1),AP=
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=
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=
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,
AC=
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=
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=3
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,
∴tan∠PCA=
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=
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=
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.
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(3)設(shè)點(diǎn)M(a,b)(a<0,b>0),分兩種情況討論:
①當(dāng)NE=2DE時(shí),NE=6,即N(-6,6),已知D(-6,-3),則有:
直線MN的斜率:k
1=
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,直線MD的斜率:k
2=
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;
由于MN⊥DM,則k
1•k
2=
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=-1,
整理得:a
2+b
2+12a-3b+18=0…(△),
由拋物線的解析式得:
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a
2+4a+9=b,
整理得:a
2+12a-3b+27=0…(□);
(△)-(□)得:b
2=9,即b=3(負(fù)值舍去),
將b=3代入(□)得:a=-6+3
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,a=-6-3
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,
故點(diǎn)M(-6+3
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,3)或(-6-3
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,3);
②當(dāng)2NE=DE時(shí),NE=
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,即N(-6,
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),已知D(-6,-3),
則有:直線MN的斜率:k
1=
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,直線DM的斜率:k
2=
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;
由題意得:k
1•k
2=
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=-1;
整理得:a
2+b
2+
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b+12a+
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=0,
而a
2+12a-3b+27=0;兩式相減,得:2b
2+9b+9=0,
解得b=-2,b=-
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,(均不符合題意,舍去);
綜上可知:存在符合條件的M點(diǎn),且坐標(biāo)為:M(-6+3
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,3)或(-6-3
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,3).
分析:(1)根據(jù)已知的拋物線解析式,可求得頂點(diǎn)D的坐標(biāo),即可求得DE、OE的長(zhǎng),根據(jù)AE2=3DE,可求出AE的值,進(jìn)而可得到點(diǎn)A的坐標(biāo),然后將其代入拋物線的解析式中,即可求得待定系數(shù)a的值,從而確定該拋物線的解析式.
(2)設(shè)出點(diǎn)P的縱坐標(biāo)為P(-6,a),根據(jù)勾股定理求出a的值,再求出AP、AC的值,從而得到tan∠PCA的值;
(3)此題比較復(fù)雜,需要分兩種情況考慮:
①NE=2DE,此時(shí)N(-6,6),可設(shè)出點(diǎn)M的坐標(biāo),然后分別表示出直線MN、直線MD的斜率,若兩條直線互相垂直,那么它們的斜率的積為-1,可據(jù)此得到關(guān)于M點(diǎn)橫、縱坐標(biāo)的關(guān)系式,聯(lián)立拋物線的解析式即可得到點(diǎn)M的坐標(biāo);
②2NE=DE,方法同①.
點(diǎn)評(píng):本題考查的是二次函數(shù)的綜合題,涉及到二次函數(shù)解析式的確定、直角三角形的判定和性質(zhì)、圓周角定理、直線與圓的位置關(guān)系、互相垂直兩直線的斜率關(guān)系等重要知識(shí),綜合性強(qiáng),難度很大.