经过点（0,√2）且斜率为k的直线l的方程为,
y - 2^(1/2) = kx,
y = 2^(1/2) + kx.
将上式带入x^2 + y^2 = 1,得
x^2 + [2^(1/2) + kx]^2 = 1,
(1+k^2)x^2 + 2k2^(1/2)x + 1 = 0
设P,Q的坐标分别为(u,2^(1/2) + ku)和(v,2^(1/2)+kv).
则,向量OP+向量OQ = [u+v,8^(1/2)+k(u+v)]
而由u,v为(1+k^2)x^2 + 2k2^(1/2)x + 1 = 0的2个实根,有
u+v = -2k2^(1/2)/[1+k^2]
向量OP+向量OQ = [u+v,8^(1/2)+k(u+v)]
// [-k,1]
又,
A的坐标为(1,0)
B的坐标为(0,1)
向量AB = [-1,1].
要使 向量OP+向量OQ 与向量AB共线,
只有,[-1,1] // [-k,1]
k = 1.