高中圆锥曲线,压轴题,可能有点难,能做多少是多少,

(1)曲线E的方程为x^2=4y
(2)解析：∵抛物线x^2=4y,A(2,1)
将直线l：y=kx+1代入抛物线得x^2-4kx-4=0
X1=2k-2√(k^2+1),X2=2k+2√(k^2+1),
Y1=2k^2-2k√(k^2+1)+1,Y2=2k^2+2k√(k^2+1)+1,
∴B(x1,y1),C(x2,y2)
AB方程：y=(1-y1)/(2-x1)*(x-2)+1
2-x1=2-2k+2√(k^2+1),1-y1=-2k^2+2k√(k^2+1)
(2-x1)/(1-y1)=1/k*[1/(√(k^2+1)-k)+1]
令(1-y1)/(2-x1)*(x-2)+1=-1
解得xs=2-2(2-x1)/(1-y1)=2-2/k*[1/(√(k^2+1)-k)+1]

AC方程：y=(1-y2)/(2-x2)*(x-2)+1
2-x2=2-2k-2√(k^2+1)；1-y2=-2k^2-2k√(k^2+1)
(2-x2)/(1-y2)=1/k*[1-1/(√(k^2+1)+k)]
令(1-y2)/(2-x2)*(x-2)+1=-1
解得xt=2-2(2-x2)/(1-y2)=2-2/k*[1-1/(√(k^2+1)+k)]
圆心横坐标：(xs+xt)/2=-2/k
圆半径：|xt-xs|/2=2√(k^2+1)/k
此圆方程：[x+2/k]^2+(y+1)^2=4(k^2+1)/k^2
(x^2+4x)/k^2+(y+1)^2-4=0
令x^2+4x=0==>x=0或x=-4
此时y=-3或y=1
∴当k≠0时,此圆必过定点(0,1)抛物线焦点,(0,-3)