(1)作BD⊥OA于点D,
∴∠ADB=90°,
∴在Rt△ABD中,由勾股定理得
AD2=AB2-BD2
∵B(3,4),
∴OD=3,BD=4.
∵AB=5,
∴AD2=25-16,
∴AD=3,
∴tan∠BAD=
.
(2)∵AD=3,OD=3,
∴OA=6,
∴A(6,0),O(0,0)
∴
∴
∴抛物线的解析式为:y=
x2?
x
∴y=
(x3)2?4,
∴M(3,-4).
(3)∵M(3,-4),B(3,4),
∴OB=OM,
∵BD⊥OA,OD=AD,
∴OB=AB=5,
∴OM=5.
△ABO∽△MQO时,
=
,
∴
=
,
∴OQ=
,
∴Q(
,0)
△ABO∽△QMO时,
=
,
∴
=
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