设动点M(x,y)满足sqr((x-2)^2+(y-3)^2)+sqr((x+4)^2+(y+5)^2)=k

(2,3)和(-4,-5)距离是
√(2+4)^2+(3+5)^2 = 10
k=12时12>10所以到两定点距离和为定值的的点轨迹是椭圆
当k=10时,则轨迹就是(2,3)和(-4,-5)这两点之间的线段
方程是
(y+5)/(x+4) = (3+5)/(2+4)
(y+5)/(x+4) = 4/3
4x+16=3y+15
4x-3y+1=0(-4