根据B1的坐标,易求得直线OB1的解析式为:y=-x; ∵OB1A1是等边三角形,且B1(-,-), ∴OA1=1,A1(0,-1); ∵直线OB1∥A1B2,又直线A1B2过点A1(0,-1), ∴直线A1B2的解析式为y=-x-1,联立抛物线的解析式,得: , 解得:(x>0); 故B2(,-2),A1A2=2,A2(0,-3); 同理可求得B3(,-),A2A3=3,A3(0,-6); … 依此类推,当A100时,A99A100=100, 点A100纵坐标的绝对值=1+2+3+…+100=5050, 故A100(0,-5050). |