∵“过点A（0,－b）和B(a,0)的直线与原点的距离为根3/2”,∴a·b = (√3/2)·√(a^2 + b^2),∵e = √6/3 = c/a ,联合解得：a = √3 ,b = 1 ,c = √2,椭圆方程：(x^2/3) + y^2 = 1.
假设C(x1,y1) ,D(x2,y2),
要使“以CD为直径的圆过点E”,则CE⊥DE,∴CE、DE斜率乘积 = -1
即：[y1/(x1 + 1)]·[y2/(x2 + 1)] = -1 ,即(y1·y2)/[(x1·x2) + (x1 + x2) + 1] = -1.T式
将y = kx + 2代入椭圆方程得：
x^2 + 3(kx + 2)^2 = 3
或3y^2 + [(y - 2)^2/(k^2)] = 3
分别整理得：(1 + 3k^2)x^2 + 12kx + 9 = 0
以及：[3 + (1/k^2)]y^2 -(4y/k^2) + [(4/k^2) - 3] = 0
根据韦达定理：x1x2 = 9/(1 + 3k^2) ,x1 + x2 = -12k/(1 + 3k^2),y1y2 = (4 - 3k^2)/(1 + 3k^2)
∴(x1·x2) + (x1 + x2) + 1 = (10 - 12k + 3k^2)/(1 + 3k^2),代入T式：
(4 - 3k^2)/(10 - 12k + 3k^2) = -1
∴6k^2 - 12k + 6 = 0,解得k = 1,直线为：y = x + 2
因此,存在k = 1使得 以CD为直径的圆过点E