若3x^2+2y^2=6x,则x^2+y^2的值域是多少
问题描述：
若3x^2+2y^2=6x,则x^2+y^2的值域是多少
答
3x^2+2y^2=6x --->3(x^2-2x)^2+2y^2=0 --->3(x-1)^2+2y^2=3 --->(x-1)^2+y^2/(3/2)=1.(*) --->x=1+cost; y=√(3/2)sint 因此x^2+y^2=(1+cost)^2+3/2*(sint)^2 =[1+2cost+(cost)^2]+3/2[1-(cost)^2] =-1/2*(cost)^2+2...