Question

求到点P（5,0）的距离和它到直线x=9/5的距离的比等于5/3的点的轨迹方程

Answer 1

设这个点为M(x,y)

PM=√[(x-5)^2+y^2]

到直线的距离d=|x-9/5|

PM/d=5/3

3√[(x-5)^2+y^2]=5|x-9/5| 两边同时平方

9[(x-5)^2+y^2]=(5x-9)^2

9x^2-90x+225+9y^2=25x^2-225x+81

16x^2-9y^2=144

x^2/9-y^2/16=1

轨迹方程为：x^2/9-y^2/16=1