Question

设P(X,Y)是曲线x^2+(y+4)^2=4上任意一点.则（x-1）^2+(y-1)^2的最大值是？

Answer 1

设x=2sint
y=2cost-4

则有
（x-1）^2+(y-1)^2
=(2sint-1)^2+(2cost-4-1)^2
=4sint^2-4sint+1+4cost^2-20cost+25
=-4sint-20cost+30
=-4(sint+5cost)+30
因为√(1^2+5^2)=√26
所以原式=-4√26(sint*1/√26+cost*5/√26)+30
令1/√26=cosA
则sinA=√(1-cos^2A)=√(1-1/26)=√25/26=5√26
所以原式=-4√26(sintcosA+costsinA)+30
=-4√26sin(t+A)+30
当sin(t+A)=-1时，（x-1）^2+(y-1)^2有最大值30+4√26