令x=3cosx,y=2sinx则求(3cosx,2sinx)与(1,0)的最短距离,由两点距离公式得,d^2=(3cosx-1)^2+(2sinx)^2 =4+1+5(cosx)^2-6cosx =5+5(cosx-3/5)^2-9/5 =16/5+5(cosx-3/5)^2当5(cosx-3/5)^2=0时,取最小.即d^2=16/5...