BQ=1×t=t,OP=1×t=t.
則OQ=6-t.
故y=
1 |
2 |
1 |
2 |
1 |
2 |
(2)①若△POQ∽△AOB時(shí),
OQ |
OB |
OP |
OA |
6?t |
6 |
t |
12 |
即12-2t=t,
解得:t=4.
②若△POQ∽△BOA時(shí),
OQ |
OA |
OP |
OB |
6?t |
12 |
t |
6 |
即6-t=2t,
解得:t=2.
∵0<t<6,
∴t=4和t=2均符合題意,
故當(dāng)t=4或t=2時(shí),△POQ與△AOB相似.
1 |
2 |
1 |
2 |
1 |
2 |
OQ |
OB |
OP |
OA |
6?t |
6 |
t |
12 |
OQ |
OA |
OP |
OB |
6?t |
12 |
t |
6 |