Center of mass of triangle created by parabola and two lines The parabola $y^2 = 3x$ is given.
Two perpendicular straight lines are drawn from the origin point, intersecting the parabola in points P and Q. 
Find equation (in cartesian form) of the set of centers of mass of all triangles OPQ (where O is the origin point).
Please help, i managed to notice, that one line is $ax$, the second is $-\frac{1}{a}x$. Also, the centers of mass move on the shape of another parabola as we rotate the lines. 

Step 1: calculate points $P$ and $Q$. You already have the equation for those lines, make them intersect with the parabola. Assume $a>0$. Note that $x>0$
$$ax=\sqrt{3x}$$ yields $$x=\frac{3}{a^2}$$ and $$y=\frac{3}{a}$$
The second equation is $y=-x/a$. You square it and get $$x=3a^2$$ and $$y=-3a$$
Step 2: The center of mass is at the average position of the vertices $(0,0)$, $(3/a^2,3/a)$, $(3a^2,-3a)$
