Centroids of triangles in parabola

Question

Two perpendicular lines intersect at $O = (0, 0)$. One of the lines intersect parabola $y^2 = 3x$ at some point $G$, the other one intersects this parabola at $H$. Find equation of the set of all centroids of triangles $OGH$. $G$ and $H$ are different from $O$.

Answer 1

A line passing through $O$ has equation $r:y=mx$ and intersects the parabola $y^2=3x$ at point $O$ and $G\left(\dfrac{3}{m^2};\;\dfrac{3}{m}\right)$

The perpendicular has equation $s:y=-\dfrac{1}{m}\,x$ and intersects the parabola at $O$ and at point $H(3m^2;\;-3m)$

Therefore the centroid of triangle $OGH$ has coordinates $P\left(\dfrac{x_O+x_G+x_H}{3};\;\dfrac{y_o+y_G+y_H}{3}\right)$

that is $P\left(m^2+\dfrac{1}{m^2};\;\dfrac{1}{m}-m\right)$

The parametric equations of the locus are

$ \left\{ \begin{array}{l} x=m^2+\dfrac{1}{m^2} \\ y=\dfrac{1}{m}-m \\ \end{array} \right. $

$y^2=m^2+\dfrac{1}{m^2}-2$

The equation of the locus is

$y^2=x-2$

Hope this is useful

Answer 2

So we have a line $y=kx$ which intersect a parabola at $G({3\over k^2}, {3\over k})$ and a line $y=-{x\over k}$ which intersect a parabola at $H(3k^2, -3k)$.

So the center of gravity is $$T({1\over k^2}+k^2, {1\over k}-k)$$

So $$y^2 = ({1\over k}-k)^2 = {1\over k^2}-2+k^2 = x-2$$