【答案】
分析:(1)根據(jù)拋物線的解析式,利用對(duì)稱軸公式,可直接求出其對(duì)稱軸.
(2)令x=0,可求出C點(diǎn)坐標(biāo),由BC∥x軸可知B,C關(guān)于拋物線的對(duì)稱軸對(duì)稱,可求出B點(diǎn)坐標(biāo),根據(jù)AC=BC可求出A點(diǎn)坐標(biāo).
(3)分三種情況討論:
①以AB為腰且頂角為∠A,先求出AB的值,再利用等腰三角形的性質(zhì)結(jié)合勾股定理求出P
1N的長(zhǎng),即可求出P
1的坐標(biāo);
②以AB為腰且頂角為角B,根據(jù)MN的長(zhǎng)和MP
2的長(zhǎng),求出P
2的縱坐標(biāo),已知其橫坐標(biāo),可得其坐標(biāo);
③以AB為底,頂角為角P時(shí),依據(jù)Rt△P
3CK∽R(shí)t△BAQ即可求出OK和P
3K的長(zhǎng),可得P
3坐標(biāo).
解答:
解:(1)拋物線的對(duì)稱軸x=-

=
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;(2分)
(2)由拋物線y=ax
2-5ax+4可知C(0,4),對(duì)稱軸x=-

=
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,
∴BC=5,B(5,4),又AC=BC=5,OC=4,
在Rt△AOC中,由勾股定理,得AO=3,
∴A(-3,0)B(5,4)C(0,4)(5分)
把點(diǎn)A坐標(biāo)代入y=ax
2-5ax+4中,
解得a=-

,(6)
∴y=

x
2+

x+4.(7分)
(3)存在符合條件的點(diǎn)P共有3個(gè).以下分三類情形探索.
設(shè)拋物線對(duì)稱軸與x軸交于N,與CB交于M.
過(guò)點(diǎn)B作BQ⊥x軸于Q,
易得BQ=4,AQ=8,AN=5.5,BM=

.
①以AB為腰且頂角為角A的△PAB有1個(gè):△P
1AB.
∴AB
2=AQ
2+BQ
2=8
2+4
2=80(8分)
在Rt△ANP
1中,P
1N=
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=
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=
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=

,
∴P
1(

,-
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).(9分)
②以AB為腰且頂角為角B的△PAB有1個(gè):△P
2AB.
在Rt△BMP
2中MP
2=

=
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=
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=

,(10分)
∴P
2=(

,
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).(11分)
③以AB為底,頂角為角P的△PAB有1個(gè),即△P
3AB.
畫(huà)AB的垂直平分線交拋物線對(duì)稱軸于P
3,此時(shí)平分線必過(guò)等腰△ABC的頂點(diǎn)C.
過(guò)點(diǎn)P
3作P
3K垂直y軸,垂足為K,
∵∠CP
3K=∠ABQ,∠CKP
3=∠AQB,
∴Rt
△P
3CK∽R(shí)t
△BAQ.
∴

=
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=
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.
∵P
3K=2.5
∴CK=5于是OK=1,(13分)
∴P
3(2.5,-1).(14分)
點(diǎn)評(píng):此題考查了用對(duì)稱軸公式求函數(shù)對(duì)稱軸方程,用待定系數(shù)法求函數(shù)解析式等基礎(chǔ)知識(shí),還結(jié)合等腰三角形的性質(zhì)考查了點(diǎn)的存在性問(wèn)題,有一定的開(kāi)放性.