SOLUTION: find the equation of the path of a point that moves so that its distance from (4,0) is twice its distance from the line x=1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the equation of the path of a point that moves so that its distance from (4,0) is twice its distance from the line x=1      Log On


   

Question 956781: find the equation of the path of a point that moves so that its distance from (4,0) is twice its distance from the line x=1
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The distance from the path to (4,0) is,
D%5B1%5D%5E2=%28x-4%29%5E2%2B%28y-0%29%5E2
I'm assuming that by distance from x=1 that you mean the perpendicular distance to x=1, that is, having no y component, only x.
The distance from the path to x=1 is,
D%5B2%5D%5E2=%28x-1%29%5E2%2B%281-1%29%5E2
.
.
.
2D%5B1%5D=D%5B2%5D
4D%5B1%5D%5E2=D%5B2%5D%5E2
4%28x-4%29%5E2%2B4y%5E2=%28x-1%29%5E2
4%28x%5E2-8x%2B16%29%2B4y%5E2=x%5E2-2x%2B1
4x%5E2-32x%2B64%2B4y%5E2=x%5E2-2x%2B1
3x%5E2-30x%2B63%2B4y%5E2=0
3%28x%5E2-10x%2B21%29%2B4y%5E2=0
3%28x%5E2-10x%2B25%29%2B21%2B4y%5E2=3%284%29
3%28x-5%29%5E2%2B4y%5E2=12
highlight%28%28x-5%29%5E2%2F4%2By%5E2%2F3=1%29
The solution is an ellipse centered at (5,0).