1.(1)直線Y=(1/2)X+2與X軸交于A(-4,0),與Y軸交于D(0,2),則:OD=2.
∵PB垂直X軸,OD·PB=8,即2*PB=8.
∴PB=4,把Y=4代入y=(1/2)x+2,得4=(1/2)x+2,x=4.故點P為(4,4).
把點P(4,4)的坐標(biāo)代入y=k/x,得k=4*4=16,反比例函數(shù)解析式為y=16/x.
(2)當(dāng)點Q與A重合時,⊿PBQ∽⊿DOA,此時Q為(-4,0);
當(dāng)點Q為(12,0)時,同理可知:⊿PBQ∽⊿DOA;
當(dāng)點Q為(2,0)時,BQ/PB=PB/BA=1/2,又∠PBQ=∠ABP=90°,⊿PBQ∽⊿AOD;
當(dāng)點Q為(6,0)時,同進可知:⊿PBQ∽⊿AOD.
2.(1)作DH垂直X軸于H,則:⊿BDH∽⊿BAO.
∴BH/BO=BD/BA,BH/t=1/2,BH=0.5t,OH=0.5t;
同理相似可求:DH=AO/2=3.故點D為(0.5t,3).
作CM垂直X軸于M,易知⊿AOB∽⊿BMC,AB/BC=AO/BM,2/1=6/BM,BM=3,OM=t+3;
AB/BC=OB/MC,2/1=t/MC,MC=0.5t.故點C為(t+3,0.5t).
(2)頂點為B(t,0),可設(shè)拋物線為:y=a(x-t)².
圖象過點C(t+3,0.5t),則:0.5t=a(t+3-t)²=9a,a=(1/18)t.
故拋物線解析式為:y=(1/18)t*(x-t)²=(t/18)x²-(t²/9)x+t³/18.
(3)AB/BC=2,AB=2BC;又∠AOB=∠ABC=90°.
則當(dāng)AO/OB=2或OB/AO=2時,以A,B,C為頂點的三角形與△AOB相似.
①當(dāng)AO/OB=2時,OB=AO/2=3,即t=3,拋物線解析式為:y=(1/6)x²-x+3/2;
②當(dāng)OB/AO=2時,OB=2AO=12,即t=12,拋物線為:y=(2/3)x²-(48/3)x+96.