【答案】
分析:(1)拋物線解析式中有兩個(gè)待定系數(shù)a,c,根據(jù)直線AC解析式求點(diǎn)A、C坐標(biāo),代入拋物線解析式即可;
(2)分析不難發(fā)現(xiàn),△ABP的直角頂點(diǎn)只可能是P,根據(jù)已知條件可證AC
2+BC
2=AB
2,故點(diǎn)C滿(mǎn)足題意,根據(jù)拋物線的對(duì)稱(chēng)性,點(diǎn)C關(guān)于拋物線對(duì)稱(chēng)軸的對(duì)稱(chēng)點(diǎn)也符合題意;
(3)由于B,F(xiàn)是定點(diǎn),BF的長(zhǎng)一定,實(shí)際上就是求BM+FM最小,找出點(diǎn)B關(guān)于直線AC的對(duì)稱(chēng)點(diǎn)B',連接B'F,交AC于點(diǎn)M,點(diǎn)M即為所求,由(2)可知,BC⊥AC,延長(zhǎng)BC到B',使BC=B'C,利用中位線的性質(zhì)可得B'的坐標(biāo),從而可求直線B'F的解析式,再與直線AC的解析式聯(lián)立,可求M點(diǎn)坐標(biāo).
解答:
解:(1)∵直線y=-
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x-
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與x軸交于點(diǎn)A,與y軸交于點(diǎn)C
∴點(diǎn)A(-1,0),C(0,-

)
∵點(diǎn)A,C都在拋物線上,
∴
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∴
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∴拋物線的解析式為y=

x
2-
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
x-
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∴頂點(diǎn)F(1,-

).
(2)存在:
p
1(0,-
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),p
2(2,-
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).
(3)存在
理由:
解法一:
延長(zhǎng)BC到點(diǎn)B′,使B′C=BC,連接B′F交直線AC于點(diǎn)M,則點(diǎn)M就是所求的點(diǎn),
∵過(guò)點(diǎn)B′作B′H⊥AB于點(diǎn)H,
∵B點(diǎn)在拋物線y=

x
2-
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
x-

上,
∴B(3,0),
在Rt△BOC中,tan∠OBC=

∴∠OBC=30°,BC=2

在Rt△B′BH中,B′H=
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BB′=2

BH=
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B′H=6,∴OH=3,
∴B′(-3,-2

).
設(shè)直線B′F的解析式為y=kx+b,
∴
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,
解得
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,
∴y=

.
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,
解得
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,
∴M(

)
∴在直線AC上存在點(diǎn)M,使得△MBF的周長(zhǎng)最小,此時(shí)M(

).
解法二:
過(guò)點(diǎn)F作AC的垂線交y軸于點(diǎn)H,則點(diǎn)H為點(diǎn)F關(guān)于直線AC的對(duì)稱(chēng)點(diǎn),連接BH交AC于點(diǎn)M,則點(diǎn)M
即為所求.
過(guò)點(diǎn)F作FG⊥y軸于點(diǎn)G,則OB∥FG,BC∥FH,
∴∠BOC=∠FGH=90°,∠BCO=∠FHG
∴∠HFG=∠CBO
同方法一可求得B(3,0)
在Rt△BOC中,tan∠OBC=


∴∠OBC=30°,可求得GH=GC=

∴GF為線段CH的垂直平分線,可證得△CFH為等邊三角形
∴AC垂直平分FH
即點(diǎn)H為點(diǎn)F關(guān)于AC對(duì)稱(chēng)點(diǎn),
∴H(0,-

)
設(shè)直線BH的解析式為y=kx+b,由題意得,

,
解得

,
∴y=

,
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,
解得
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,
∴M(

),
∴在直線AC上存在點(diǎn)M,使得△MBF的周長(zhǎng)最小,此時(shí)M(

).
點(diǎn)評(píng):考查代數(shù)幾何的綜合運(yùn)用能力,體現(xiàn)數(shù)學(xué)知識(shí)的內(nèi)在聯(lián)系和不可分割的特點(diǎn).