已知点p(x,y)到原点的距离为1,则x+y-2/x-y+2的最大值是多少

Question

已知点p(x,y)到原点的距离为1,则x+y-2/x-y+2的最大值是多少

Answer 1

p在单位圆上,可设p(cost,sint)
则f=(x+y-2)/(x-y+2)=(cost+sint-2)/(cost-sint+2)
cost+sint-2=fcost-fsint+2f
(1-f)cost+(1+f)sint=2+2f
√[(1-f)^2+(1+f)^2]sin(t+p)=2(1+f),这里tanp=(1-f)/(1+f)
因此有4(1+f)^2