1.(1)直线Y=(1/2)X+2与X轴交于A(-4,0),与Y轴交于D(0,2),则:OD=2.
∵PB垂直X轴,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.故点P为(4,4).
把点P(4,4)的坐标代入y=k/x,得k=4*4=16,反比例函数解析式为y=16/x.
(2)当点Q与A重合时,⊿PBQ∽⊿DOA,此时Q为(-4,0);
当点Q为(12,0)时,同理可知:⊿PBQ∽⊿DOA;
当点Q为(2,0)时,BQ/PB=PB/BA=1/2,又∠PBQ=∠ABP=90°,⊿PBQ∽⊿AOD;
当点Q为(6,0)时,同进可知:⊿PBQ∽⊿AOD.
2.(1)作DH垂直X轴于H,则:⊿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.故点D为(0.5t,3).
作CM垂直X轴于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.故点C为(t+3,0.5t).
(2)顶点为B(t,0),可设抛物线为:y=a(x-t)².
图象过点C(t+3,0.5t),则:0.5t=a(t+3-t)²=9a,a=(1/18)t.
故抛物线解析式为: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°.
则当AO/OB=2或OB/AO=2时,以A,B,C为顶点的三角形与△AOB相似.
①当AO/OB=2时,OB=AO/2=3,即t=3,抛物线解析式为:y=(1/6)x²-x+3/2;
②当OB/AO=2时,OB=2AO=12,即t=12,抛物线为:y=(2/3)x²-(48/3)x+96.