题目

实数xy满足x2+y2+2x-4y+1=0,则x2+y2-2x+1的最小值为

提问时间：2020-06-25

答案

x^2+y^2+2x-4y+1=0
(x+1)^2+(y-2)^2=2^2
x=-1+2cosx,y=2+2sinx
y-x
=3+2(sinx-cosx)
=3+2√2(sinx/√2-cosx/√2)
=3+2√2(sinx*cos45°-cosx*sin45°)
=3+2√2sin(x-45°)
-1≤sin(x-45°)≤1
3-2√2≤y-x≤3+2√2
已知x^2+y^2+2x-4y+1=0,则
x^2+y^2+2x-4y+1+4y-4x=4y-4x
x^2+y^2-2x+1=4(y-x)
4(3-2√2)≤x^2+y^2-2x+1≤4(3+2√2)
x^2+y^2-2x+1的最小值=4(3-2√2)