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(1)A的坐标为:(t+1,t^2)而:Y=x^2-2x+1=(x-1)^2,所以B的坐标为:(1,0)显然,A的坐标满足Y=(x-1)^2所以:点A在抛物线Y=x^2-2x+1上(2)将(1,0)代入Y=a(x-t-1)^2+t^2,at^2+t^2=0(a+1)t^2=0因t不等于0a+1=0a=-1解-(x-t-1)^2+t^2=0得:x=1 及x=2t+1这条抛物线与X轴的两个交点为:B(1,0),C(2t+1,0)BC=2tAB=((t+1-1)^2+t^4)^(1/2)=t(t^2+1)^(1/2)AC=((t+1-2t-1)^2+t^4)=t(t^2+1)^(1/2)如:... |
