x=α +(t^2+1)b=(1-2(t^2+1),2+t^2+1)
y=(-k,-2k)+ (1/t)*b=( -k-2/t, -2k+1/t)
(1)x与y垂直
(-1-2t^2)(-k-2/t)+(t^2+3)(-2k+1/t)=0
k+2/t+2kt^2+4t-2kt^2+t-6k+3/t=0
1/t+t-k=0
k=t+1/t≥2√t(1/t)=2
k的最小值=2
(2)x∥y
[-2t^2-1]/[-k-2/t]=[t^2+3]/[-2k+1/t)]
(2t^2+1)(-2k+1/t)=(t^2+3)(k+2/t)
t^2k+k+1/t=0
(1+t^2)k=-1/t
k=(-1/t)/(1+t^2)
k,t为正实数
k=(-1/t)/(1+t^2)<0
与已知矛盾
所以,不存在k,t使x∥y