(1)设抛物线的解析式为y=kx2+a ∵点D(2a,2a)在抛物线上 4a2k+a = 2a ∴k = ∴抛物线的解析式为y= x2+a (2)设抛物线上一点P(x,y),过P作PH⊥x轴,PG⊥y轴,在Rt△GDP中, 由勾股定理得:PD2=DG2+PG2=(y–2a)2+x2 =y2 – 4ay+4a2+x2 ∵y= x2+a ∴x2 =" 4a" ´ (y– a)=" 4ay–" 4a2 ∴PD 2= y2– 4ay+4a2 +4ay– 4a2= y2 =PH2 ∴PD = PH (3)过B点BE ⊥ x轴,AF⊥x轴
由(2)的结论:BE=DB AF=DA ∵DA=2DB ∴AF=2BE ∴AO = 2BO ∴B是OA的中点∴C是OD的中点 连结BC ∴BC= = =" BE" = DB 过B作BR⊥y轴 ∵BR⊥CD ∴CR=DR,OR=" a" + = ∴B点的纵坐标是,又点B在抛物线上 ∴ = x2+a ∴x2 =2a2,∵x>0 ∴x = a,∴B (a, ) AO = 2OB, ∴S△ABD=S△OBD = 4 所以,´2a´a= 4 ∴a2= 4 ∵a>0 ∴a = 2 |