题目内容

已知点O(0,0),A(0,b),B(a,a3).若△AOB为直角三角形,则必有(  )
A.b=a3B.(b-a3)(b-a3-
1
a
)=0
C.|b-a3|+|b-a3-
1
a
|=0
D.b=a3+
1
a
AB
=(a,a3-b),
OA
=(0,b),
OB
=(a,a3),且ab≠0
OA
OB
,则
OA
OB
=ba3=0,∴a=0或b=0,但是ab≠0,应舍去;
OA
AB
,则
OA
AB
=b(a3-b)=0,∵b≠0,∴b=a3≠0;
OB
AB
,则
OB
AB
=a2+a3(a3-b)=0,得1+a4-ab=0,即 b-a3-
1
a
=0
综上可知:△OAB为直角三角形,则必有(b-a3)(b-a3-
1
a
)=0
故选:B.
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