已知向量a=（根号3,1）,向量b=（1/2,根号3/2）,且存在实数k和t,使得x=a+(t^2-3)b,y=-ka+tb,且x垂直y试求﹙k＋t^2﹚／t的最小值

X=a+(t²-3)b=(√3+(t²-3)/2,-1+(√3)(t²-3)/2)；Y=(-k√3 +t/2,k+(√3)t/2)；∵X⊥Y,∴X•Y=[√3+(t²-3)/2][-k√3 +t/2]+[-1+(√3)(t²-3)/2][k+(√3)t/2]=[-3k-(√3)k(t²-3)/2+(√3)t/2+t(t²-3)/4]+[-k+(√3)k(t²-3)/2-(√3)t/2+3t(t²-3)/4]=-4k+4t(t²-3)/4=0故k=t(t²-3)/4,∴u=(k+t²)/t=[t(t²-3)/4+t²]/t=(t³+4t²-3t)/4t=(t²+4t-3)/4=(1/4)[(t+2)²-7]≧-7/4当且仅仅当t=-2(此时k=-1/2)时等号成立.即当t=-2,k=-1/2时,(k+t²)/t获得最小值-7/4.