(1)设方程为x^2/a^2+y^2/b^2=1,代入A(0,2)、B(1/2,√2 )得a,b
椭圆方程为2x^2+y^2/4=1
(2)设直线为y=k(x-1) M(X1,Y1),N(X2,Y2).(向量EM)·(向量EN)=(X1-1)(X2-1)+Y1Y2=X1X2-(X1+X2)+1+K(X1-1)K(X2-1)=(K^2+1)X1X2-(K^2+1)(X1+X2)+1+K^2,
X1+X2,X1X2可由直线与椭圆联立方程,消 y得(8+k^2)x^2-2k^2x+(k^2-4)=0
由有两个根得(-2k^2)^2-4(8+k^2)(k^2-4)>0,得k^2