Browse Questions

# Find the centroid of the triangle formed by the feet of the three normal lies on the axis of parabola?

$\begin{array}{1 1}(a)\;(\large\frac{2h-4a}{3},\normalsize 0)\\(b)\;(\large\frac{2h-2a}{3},\normalsize 1)\\(c)\;(\large\frac{3h-4a}{3},\normalsize 3)\\(d)\;(\large\frac{2h+4a}{3},\normalsize 1)\end{array}$

If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ be vertices of $\Delta ABC$ then its centroid is
$\big(\large\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$
Here $y_1+y_2+y_3=0$
$y_1=-2am_1,y_2=-2am_2,y_3=-2am_3$
$y_1+y_2+y_3=-2a(m_1+m_2+m_3)$
$m_1+m_2+m_3=0$
So centroid of $\Delta ABC\Rightarrow (\large\frac{x_1+x_2+x_3}{3},$$0) Now \large\frac{x_1+x_2+x_3}{3}=\frac{a}{3}$$(m_1^2+m_2^2+m_3^2)$
$\Rightarrow \large\frac{a}{3}$$[(m_1+m_2+m_3)^2-2(m_1m_2+m_2m_3+m_3m_1)] \Rightarrow \large\frac{a}{3}$$[0-2\times \big(\large\frac{2a-h}{a}\big)]$
$\Rightarrow \large\frac{2h-4a}{3}$
Hence centroid of $\Delta ABC$ is $\big(\large\frac{2h-4a}{3}$$,0)$
Hence (a) is the correct answer.