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.

  • $\begingroup$ 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

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 $(1)$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.