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Show that the locus of the centroids of equilateral triangles inscribed in the parabola $y^2=4ax$ is the parabola $9y^2-4ax+32a^2=0.$

I tried to solve it.I took three coordinates of the equilateral triangle as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$

let the coordinates of the centroid be $(h,k)$.

$h=\frac{x_1+x_2+x_3}{3},k=\frac{y_1+y_2+y_3}{3}$

$h=\frac{y^2_1+y^2_2+y^2_3}{12a}$

But eliminating $y_1,y_2,y_3$ is difficult.Is my method correct or is there any other better method.Please help me.

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If you want to try yourself, translate into algebra the fact the triangle of vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is equilateral. Two equations are enough since congruence is transitive: don't use distance but orthogonality. Then you need some algebraic manipulations to get the result.

Otherwise the complete solution is in Gelca,Andreescu Putnam and Beyond (2007), pp. 213-214.

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