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How to prove that the centroid of a triangle formed by three co-normal points lies on the axis of the parabola?

Note: "Co-normal points" are the feet of normals drawn from a point to the parabola.

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  • $\begingroup$ Can you define co-normal for us please? $\endgroup$ – David H Apr 12 '14 at 8:02
  • $\begingroup$ Co normal points are the feet of normals drawn from a point to the parabola $\endgroup$ – user34304 Apr 12 '14 at 8:09
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WLOG, we can consider the equation of the Parabola to be $\displaystyle y^2=4ax\ \ \ \ (1)$

From this, the eqaution of the normal at $P(am^2,-2am)$ is $y = mx – 2am – am^3\iff am^3+2a m+y-mx=0$

which is a Cubic Equation in $m$ having three roots $m_1,m_2,m_3$(say)

Using Vieta's formulas, $\displaystyle m_1+m_2+m_3=-\frac0a=0$

So, the ordinate of the centroid of the Triangle will be $\displaystyle\frac{-2a(m_1+m_2+m_3)}3=0$ hence, the centroid will lie on the $X$ axis which is evidently the axis of the Parabola $(1)$

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  • $\begingroup$ Really nice, sir!!! $\endgroup$ – user34304 Apr 13 '14 at 4:44

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