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. 

A generic line through the origin has equation $r:y=mx$ and we have $r^{\perp}:y=-\frac{1}{m}x$
The intersection of $r$ and $r^{\perp}$ with the parabola $y^2=3x$ are respectively $P\left(\frac{3}{m^2};\;\frac{3}{m}\right)$ and $Q(3m^2,\;-3m)$
The solution of the systems 
$
\left\{
\begin{array}{l}
 y^2=3x \\
 y=mx \\
\end{array}
\right.\quad\quad
$
$
\left\{
\begin{array}{l}
 y^2=3x \\
 y=-\frac{1}{m}\,x \\
\end{array}
\right.
$
Centroid $C$ of triangle $OPQ$ has coordinates which are the average of the coordinates of $(O,P,Q)$ that is $C\left(\frac{1}{m^2}+m^2,\frac{1}{m}-m\right)$
Point $C$ as $m$ varies describes a curve of parametric equations
$
\left\{
\begin{array}{l}
 x=m^2+\frac{1}{m^2} \\
 y=\frac{1}{m}-m \\
\end{array}
\right.
$
To eliminate the parameter let's square $y$
$\left(\frac{1}{m}-m \right)^2=\frac{1}{m^2}+m^2-2$
so we get the equation
$y^2=x-2$
Hope this is useful
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