(1)令-x2+2ax+1-a2=0∴由根与系数的关系知:x1*x2=a^2-1,x1+x2=2a,|AB|=|x1-x2|,而|x1-x2|^2=(x1+x2)^2-4*(x1*x2)=4,所以|AB|=22)y=-x2+2ax+1-a2=1-(x-a)^2,于是c(a,1),向量CA=(x1-a,-1)向量CB=(x2-a,-1),向量CA点乘向量CB=x1*x2-a(x1+x2)+a^2+1,根据根与系数的关系,代入x1*x2=a^2-1,x1+x2=2a可得乘积为0,因而是直角三角形3)s=(1/2)*|AB|*yd,yd是D点纵坐标的绝对值,又D(0,1-a^2)... |