Question

M(x,y)在圆x^2+y^2=1上移动，求点Q(x(x+y),y(x+y))的轨迹

Answer 1

用参数方程来表示简单些。
设 M(x,y) 圆的参数方程：
x=cost, y=sint
Q点坐标
x=cost(cost+sint)=cos^2t+cost*sint
=1/2*cos2t+1/2+1/2*sin2t
= √2/2 (√2/2*cost2t + √2/2*sin2t)+1/2
= √2/2 cost(2t-π/4)+1/2

y=sint(cost+sint)=sin^2t+sint*cost
=1/2-1/2*cos2t+1/2*sin2t
=1/2+√2/2 (√2/2*sin2t - √2/2*cos2t)
=√2/2 sin(2t-π/4)+1/2

即Q点的轨迹是以 (1/2, 1/2) 为圆心，√2/2 为半径的圆。

Answer 2

令a=x(x+y)
b=y(x+y)
x=cost
y=sint
a+b=1+2sint
cost
sintcost=(a+b-1)/2
ab=xy(x+y)^2
=(a+b-1)/2
*(x^2+y^2+2xy)
=(a+b-1)/2
*(1+a+b-1)
=(a+b-1)/2
*(a+b)
所以2ab=(a+b)(a+b-1)
2ab=(a+b)^2-a-b=a^2+b^2+2ab-a-b=0
得a^2+b^2-a-b=0
(a-1/2)^2
+(b-1/2)^2
=1/2
所以q的轨迹为(x-1/2)^2+(y-1/2)^2=1/2
这是一个圆,圆心为(1/2,1/2)
,半径为根2
/2