题目 #17581

(1)将 (−1,0), (3,0)代入y =\(ax^{2}\)+ bx + 3， A B  − + 3 = 0  = −1 得：\(\frac{a}{a} \frac{b}{3}\)，解得：\(\frac{a}{b}\)， 9 + b + 3 = 0  =\(\frac{2}{x}\)抛物线的解析式为y = −\(x^{2}\)+ 2 + 3． (2)过点作EF ⊥ x 轴于点，加图所示． E F   +  = 90,ODP + FDE = 90,\(\frac{OPD}{OPD} \frac{ODP}{FDE}\) = .\(\frac{ODP}{POD} \frac{FED}{DFE}\)在 和 中，  =  = 90    = FDE \(\frac{OPD}{DP}\) =\(\frac{ED}{FED}\)ODP  （），\(\frac{AAS}{DO}\) =, =. 抛物线的解析式\(\frac{DF}{y} \frac{OPEF}{2}\)为 = −\(x^{2}\)+ x + 3 = −(x − +，frac1)^2EF frac4DO点 的坐标为(1,0)， = = 1． D 当 = 时，−\(x^{2}\)+ 2x + 3 =1，\(\frac{y}{解得} \frac{1}{x_{1}}\): =1− （舍去），\(\frac{3}{DF}x_{2}\)= +\(\sqrt{3}\)， = OP = 1+\(\sqrt{3}\)−1，\(\frac{1}{P}\)点的坐标为(0, 3)．