aomox

(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=

4
3

(2)∵AD=3,OD=3,

∴OA=6,

∴A(6,0),O(0,0)

0=c
0=
4
9
×36?+6b+c

b=?
8
3
c=0

∴抛物线的解析式为:y=

4
9
x2?
8
3
x

∴y=

4
9
(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,

AO
MO
BO
OQ

6
5
5
OQ

∴OQ=

25
6

∴Q(

25
6
,0)

△ABO∽△QMO时,

AO
QO
BO
MO

6
QO
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