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

题目

设动点M(x,y)满足sqr((x-2)^2+(y-3)^2)+sqr((x+4)^2+(y+5)^2)=k
(1)当k=12时,点M的轨迹是哪种曲线?
(2)当k=10时,点M的轨迹是哪种曲线?并求出这条曲线的方程

答案

(2,3)和(-4,-5)距离是√(2+4)^2+(3+5)^2 = 10k=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) = 4/34x+16=3y+1...