(1)∵OA=5,AB=10,OC=12,
∴點(diǎn)B(10,5),C(12,0),
∴
,
解得
,
∴拋物線的函數(shù)表達(dá)式為y=-
x
2+3x;
(2)根據(jù)勾股定理,AC=
=
=13,
∵點(diǎn)P沿AC以每秒2個(gè)單位長(zhǎng)度的速度向點(diǎn)C運(yùn)動(dòng),點(diǎn)Q沿CO以每秒1個(gè)單位長(zhǎng)度的速度向點(diǎn)O運(yùn)動(dòng),
∴點(diǎn)P運(yùn)動(dòng)的時(shí)間為:13÷2=6.5秒,
CP=AC-AP=13-2t,CQ=t,
∵∠ACO≠90°,
∴分∠PQC=90°和∠CPQ=90°兩種情況討論:
①∠PQC=90°時(shí),cos∠ACO=
=
,
即
=
,
解得t=
,
②∠CPQ=90°時(shí),cos∠ACO=
=
,
即
=
,
解得t=
,
綜上所述,t為
秒或
秒時(shí),△PQC是直角三角形;
(3)拋物線對(duì)稱軸為直線x=-
=-
=6,
①AC是平行四邊形的邊時(shí),(i)若點(diǎn)M在對(duì)稱軸左邊,
∵OC=12,
∴點(diǎn)M的橫坐標(biāo)為:6-12=-6,
代入拋物線解析式得,y=-
×(-6)
2+3×(-6)=-27,
此時(shí)點(diǎn)M的坐標(biāo)為(-6,-27),
∵OA=5,
∴點(diǎn)N的縱坐標(biāo)為:-27-5=-32,
∴點(diǎn)N的坐標(biāo)為(6,-32);
(ii)若點(diǎn)M在對(duì)稱軸右邊,∵OC=12,
∴點(diǎn)M的橫坐標(biāo)為:6+12=18,
代入拋物線解析式得,y=-
×18
2+3×18=-27,
此時(shí)點(diǎn)M的坐標(biāo)為(18,-27),
∵OA=5,
∴點(diǎn)N的縱坐標(biāo)為:-27+5=-22,
∴點(diǎn)N的坐標(biāo)為(6,-22);
②AC是對(duì)角線時(shí),∵點(diǎn)P是AC的中點(diǎn),點(diǎn)N在對(duì)稱軸上,
∴點(diǎn)M也在拋物線對(duì)稱軸上,
∴點(diǎn)M為拋物線的頂點(diǎn),
∵y=-
x
2+3x=-
(x-12x+36)
2+9=-
(x-6)
2+9,
∴M(6,9),
∵OA=5,OC=12,點(diǎn)P在對(duì)稱軸上,
∴點(diǎn)P的坐標(biāo)為(6,
),
∴點(diǎn)N的縱坐標(biāo)為:2×
-9=-4,
∴點(diǎn)N(6,-4);
綜上所述,M(-6,-27)、N(6,-32)或M(18,-27)、N(6,-22)或M(6,9)、N(6,-4)時(shí),以M、N、A、C為頂點(diǎn)的四邊形是平行四邊形.