动圆M与两定圆F1:x^2+y^2+10x+24=0,F2:x^2+y^2-10x-24=0都外切,求动圆圆心M的轨迹方程
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/19 09:28:37
![动圆M与两定圆F1:x^2+y^2+10x+24=0,F2:x^2+y^2-10x-24=0都外切,求动圆圆心M的轨迹方程](/uploads/image/z/6087384-0-4.jpg?t=%E5%8A%A8%E5%9C%86M%E4%B8%8E%E4%B8%A4%E5%AE%9A%E5%9C%86F1%3Ax%5E2%2By%5E2%2B10x%2B24%3D0%2CF2%3Ax%5E2%2By%5E2-10x-24%3D0%E9%83%BD%E5%A4%96%E5%88%87%2C%E6%B1%82%E5%8A%A8%E5%9C%86%E5%9C%86%E5%BF%83M%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B)
动圆M与两定圆F1:x^2+y^2+10x+24=0,F2:x^2+y^2-10x-24=0都外切,求动圆圆心M的轨迹方程
动圆M与两定圆F1:x^2+y^2+10x+24=0,F2:x^2+y^2-10x-24=0都外切,求动圆圆心M的轨迹方程
动圆M与两定圆F1:x^2+y^2+10x+24=0,F2:x^2+y^2-10x-24=0都外切,求动圆圆心M的轨迹方程
F1:(x+5)^2+y^2=1,圆心为(-5,0),半径为1
F2:(x-5)^2+y^2=49,圆心为(5,0),半径为7
设动圆圆心M(x,y),半径为r,则r=MF1-1=MF2-7
即MF1+6=MF2
√[(x+5)^2+y^2]+6=√[(x-5)^2+y^2]
两边平方:20x+36+12√[(x+5)^2+y^2]=0
3√[(x+5)^2+y^2]=-5x-9
再平方:9[(x+5)^2+y^2]=25x^2+81+90x
16x^2-9y^2-144=0
即x^2/9-y^2/16=1
这是双曲线(根据r>0,-5x-9>0,知轨迹只是双曲线的一段).