x^2+y^2=2(m+3)x+2(1-4m^2)+16m^2+9=0 --->x^2+y^2-2(m+3)x+2(4m^2-1)y=-16m^2-9 --->[x-(m+3)]^2+[y+(4m^2-1)]^2=(m+3)^2+(4m^2-1)^2-16m^2-9 ..............................=16m^4-23m^2+6m+1 符合圆的方程的特征，所以圆心坐标是 x=m+3,y=-4m^2+1 消去m，由x=m+3--->m=x-3 代人y=-4(x-3)^2+1 这就是所要求的轨迹方程
x^2 + y^2 + 2(m+3)x + 2(1-4m^2) + 16m^2 + 9 = 0;
(x+(m+3))^2 + (y+(1-4m^2))^2 = r^2
r^2 = (m+3)^2 + (4m^2-1)^2 - 16m^2 -9
= 16m^4 -23m^2 + 6m+1
= 16(m^2 -1)^2 + 9(m+1/3)^2 -16>0
圆心( -m-3, 4m^2-1); 设圆心(x,y);
x = -m-3;
y = 4m^2-1;
m = -3-x,
y = 4m^2-1
= 4(x+3)^2 -1
圆心轨迹方程是抛物线.
x满足约束：
16((x+3)^2 -1)^2 + 9(x+8/3)^2 -16>0.
答：由已知方程得，[x-（t+3）]^2+[y+(1-4t^2)]^2=-7t^2+6t+1. 设圆心坐标为(m,n)，则 m=t+3,且n=-(1-4t^2)=4...详情>>
答：详情>>
