Let the coordinates of A and B be (X 1, Y 1) and (X2, Y2) respectively, and the coordinates of point P in AB be (x, y).

So there are: x 1+x2=2x, y 1+y2=2y.

Also: x12-y12 = 6, x2 2-y2 2 = 6.

Subtract two expressions: (x1+x2) (x1-x2)-(y1+y2) (y1-y2) = 0.

So the slope of AB is: k = (y1-y2)/(x1-x2) = (x1+x2)/(y1+y2) = x/y.

And: K=3.

So, x/y=3.

That is, the trajectory equation is y=x/3.