Centroids of Triangles on a Parabola

The parabola \(y=x^2\) has three points \(P_1,P_2,P_3\) on it. The lines tangent to the parabola at \(P_1, P_2, P_3\) intersect each other pairwise at \(X_1,X_2,X_3\). Let the centroids of \(\triangle P_1P_2P_3\) and \(\triangle X_1X_2X_3\) be \(G_P, G_X\) respectively. Prove that \(G_PG_X\) is parallel to the y-axis.

Note by Daniel Liu
3 years, 2 months ago

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General parametric coordinates of \(x^2 = 4ay\) is

\((2at, at^2)\)

And point of intersection of tangents from \(t_1, t_2\) is

\((a(t_1 + t_2), at_1t_2)\)

Let P,Q,R be \(t_1,t_2,t_3\) resp.

We just need to check the x-coordinate of the Centroids are same or not

x-coordinate of Centroid of PQR is \((\dfrac{2a(t_1 + t_2 + t_3)}{3})\)

x- coordinates of X,Y,Z are \(a(t_1 + t_2), a(t_2+ t_3), a(t_3 + t_1)\)

x- coordinate of Centroid of XYZ is

\(\dfrac{2a(t_1 + t_2 + t_3)}{3}\)

Hence Proved.


I m just lazy to prove point of intersection of tangents, I'll do if you want the proof.

Krishna Sharma - 3 years, 2 months ago

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