【答案】
分析:(1)由∠ACB=90°,AC=BC,OA=1,OC=4,可得A(-1,0)B(4,5),然后利用待定系數(shù)法即可求得b,c的值;
(2)由直線AB經(jīng)過點(diǎn)A(-1,0),B(4,5),即可求得直線AB的解析式,又由二次函數(shù)y=x
2-2x-3,設(shè)點(diǎn)E(t,t+1),則可得點(diǎn)F的坐標(biāo),則可求得EF的最大值,求得點(diǎn)E的坐標(biāo);
(3)①順次連接點(diǎn)E、B、F、D得四邊形EBFD,可求出點(diǎn)F的坐標(biāo)(
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,
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),點(diǎn)D的坐標(biāo)為(1,-4)由S
四邊形EBFD=S
△BEF+S
△DEF即可求得;
②過點(diǎn)E作a⊥EF交拋物線于點(diǎn)P,設(shè)點(diǎn)P(m,m
2-2m-3),可得m
2-2m-3=
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,即可求得點(diǎn)P的坐標(biāo),又由過點(diǎn)F作b⊥EF交拋物線于P
3,設(shè)P
3(n,n
2-2n-3),可得n
2-2n-2=-
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,求得點(diǎn)P的坐標(biāo),則可得使△EFP是以EF為直角邊的直角三角形的P的坐標(biāo).
解答:解:(1)由已知得:A(-1,0),B(4,5),
∵二次函數(shù)y=x
2+bx+c的圖象經(jīng)過點(diǎn)A(-1,0),B(4,5),
∴
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,
解得:b=-2,c=-3;
(2)如圖:∵直線AB經(jīng)過點(diǎn)A(-1,0),B(4,5),
∴直線AB的解析式為:y=x+1,
∵二次函數(shù)y=x
2-2x-3,
∴設(shè)點(diǎn)E(t,t+1),則F(t,t
2-2t-3),
∴EF=(t+1)-(t
2-2t-3)=-(t-
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)
2+
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,
∴當(dāng)t=
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時,EF的最大值為
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,
∴點(diǎn)E的坐標(biāo)為(
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,
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);
(3)①如圖:順次連接點(diǎn)E、B、F、D得四邊形EBFD.
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可求出點(diǎn)F的坐標(biāo)(
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,
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),點(diǎn)D的坐標(biāo)為(1,-4)
S
四邊形EBFD=S
△BEF+S
△DEF=
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×
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×(4-
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)+
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×
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×(
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-1)=
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;
②如圖:
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ⅰ)過點(diǎn)E作a⊥EF交拋物線于點(diǎn)P,設(shè)點(diǎn)P(m,m
2-2m-3)
則有:m
2-2m-3=
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,
解得:m
1=1+
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,m
2=1-
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,
∴P
1(1-
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,
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),P
2(1+
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,
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),
ⅱ)過點(diǎn)F作b⊥EF交拋物線于P
3,設(shè)P
3(n,n
2-2n-3)
則有:n
2-2n-3=-
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,
解得:n
1=
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,n
2=
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(與點(diǎn)F重合,舍去),
∴P
3(
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,-
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),
綜上所述:所有點(diǎn)P的坐標(biāo):P
1(1+
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,
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),P
2(1-
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,
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),P
3(
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,-
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)能使△EFP組成以EF為直角邊的直角三角形.
點(diǎn)評:此題考查了待定系數(shù)法求二次函數(shù)的解析式,四邊形與三角形面積問題以及直角三角形的性質(zhì)等知識.此題綜合性很強(qiáng),解題的關(guān)鍵是注意方程思想與數(shù)形結(jié)合思想的應(yīng)用.