一道解析问题有关直角距离
有关直角距离
设x1=coct，y1=sint， 则|x1-x2|+|y1-y2| =|cost-x2|+|sint-2x2+2√5| ≥|cost-x2|+|sint-2x2+2√5|/2 ≥|cost-x2-0.5sint+x2-√5| =|cost-0.5sint-√5| =|(√5/2)cos(t+a)-√5| ≥|(√5/2)-√5| =√5/2.