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1)设抛物线y=a(x-1)(x-4),过(0,-2)-2=4a, a=-1/2, 抛物线y=-(x-1)(x-4)/2=-x^2/2+5x/2-22)△OAC是直角三角形,OC=2,OA=4设P(t,-t^2/2+5t/2-2)(t≠1或4),△APM中,AMP=90°,AM=|4-t|,PM=|-t^2/2+5t/2-2|①AM=2PM, 即|4-t|=|-t^2+5t-4|,1=|t-1|t=0或2,此时P(0,-2)(这时候APM与COA重合)或P(2,1)②PM=2AM, 即4|4-t|=|-t^2+5t-4|, 4=|t-1|t=5或-3,此时P(5,-2)或P(-3,-14)综上,存在符合条件的P点,P(-3,-14),P(0 |
