Question

已知实数x、y满足（x+y）^2+y^2=1/4，则根号下（x-1）^2+(y-3)^2的最大值是多少

Answer 1

(x+y)^2+y^2=1/4
x^2+2xy+2y^2=1/4
x=x'cosu-y'sinu
y=x'sinu+y'cosu
x^2+2xy+2y^2=x'^2(cosu^2+2sinu^2)+2y'^2(sinu^2+2cosu^2)+2x'y'(cosu^2-sinu^2)
cosu=sinu=√2/2
x^2+2xy+2y^2=(3/2)x'^2+3y'^2=1/4
6x'^2+12y'^2=1
x'^2/(1/6)+y'^2/(1/12)=1
x=1,y=3 P(1,3)在x'oy'坐标系中
(x'-y')=√2,(x'+y')=3√2 x'=2√2,y'=√2
在椭圆x'^2/(1/6)+y'^2/(1/12)=1
y'=(1/2)x'交于Q
6x'^2+3x'^2=1
x'=-1/3 ,y'=-1/6
在xoy坐标系中Qx=(√2/2)(-1/3+1/6)=-√2/12
Qy=(√2/2)(-1/2)=-√2/4
√[(x-1)^2+(y-3)^2]最大=√(-√2/12-1)^2+(-√2/4-3)^2