【答案】
分析:(1)依題意設(shè)拋物線解析式為y=ax(x-4),把B(5,5)代入求得解析式.
(2)過點B作BD⊥y軸于點D,求出點C的坐標(biāo).設(shè)直線l的解析式為y=kx-4,把點B的坐標(biāo)代入求出k值之后可求出直線l的解析式.
(3)首先證明△PBQ∽△OBC根據(jù)線段比求出P
2,然后可知拋物線y=x
2-4x與直線l的交點就是滿足題意的點Q,令x
2-4x=

x-4求出P
1的坐標(biāo).然后分情況討論點P的坐標(biāo)的位置.
解答:解:(1)∵拋物線y=ax
2+bx+c經(jīng)過點(0,0),(4,0),
可設(shè)拋物線解析式為y=ax(x-4),
把B(5,5)代入,
解得a=1,
∴拋物線解析式為y=x
2-4x.(4分)
(2)過點B作BD⊥y軸于點D.
∵點B的坐標(biāo)為(5,5),

∴BD=5,OD=5.
∵tan∠OCB=

=

,
∴CD=9,
∴OC=CD-OD=4.
∴點C坐標(biāo)為(0,-4).(2分)
設(shè)直線l的解析式為y=kx-4,
把B(5,5)代入,得5=5k-4,
解得k=

.
∴直線l的解析式為y=

x-4.(2分)
(3)當(dāng)點P在線段OB上(即0<x<5時),
∵PQ∥y軸,
∴∠BPQ=∠BOC=135度.
當(dāng)

=

時,△PBQ∽△OBC.
這時,拋物線y=x
2-4x與直線l的交點就是滿足題意的點Q,
那么x
2-4x=

x-4,
解得x
1=5(舍去),x
2=

,
∴P
1(

,

);(2分)
又當(dāng)

=

時,△PQB∽△OBC.
∵PB=

(5-x),PQ=x-(x
2-4x)=5x-x
2,OC=4,OB=5

,
∴

,
整理得2x
2-15x+25=0,
解得x
1=5(舍去),x
2=

,
∴P
2(

,

).(2分)
當(dāng)點P在點O左側(cè)(即x<0=時),
∵PQ∥y軸,
∴∠BPQ=45°,△BPQ中不可能出現(xiàn)135°的角,這時以P,Q,B為頂點的三角形不可能與△OBC相似.
當(dāng)點P在點B右側(cè)(即x>5)時,
∵∠BPQ=135°,
∴符合條件的點Q即在拋物線上,同時又在直線l上;
或者即在拋物線上,同時又在Q
2,B所在直線上(Q
2為上面求得的P
2所對應(yīng)).
∵直線l(或直線Q
2B)與拋物線的交點均在0<x≤5內(nèi),而直線與拋物線交點不可能多于兩個,
∴x>5時,以P,Q,B為頂點的三角形也不可能與△OBC相似.
綜上所述,符合條件的點P的坐標(biāo)只有兩個:P
1(

,

),P
2(

,

).(2分)
點評:本題考查的是二次函數(shù)的有關(guān)知識,特別要注意的是考生需全面分析討論從而求解.