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