Find the equation of the path that moves so that its distance from (4,0) is twice its distance from the line x = 1
Take a point (u,v) on the path. Its distance from the line x=1 is |u-1|, and its
distance from (4,0) is sqrt[(u-4)^2+v^2], so:
sqrt[(u-4)^2+v^2] = 2|u-1|
now we square:
(u-4)^2+v^2 = 4(u-1)^2
which simplifies to:
3 u^2 - v^2 = 12.
Now replace u by x and v by y to get the equation with standard veriables:
3 x^2 - y^2 = 12.
RonL