The centroid of the triangle formed by joinin | vedclass

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(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.If the normal passes through a point $(h, k)

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(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.If the normal passes through a point $(h, k)$,then $k = mh - 2am - am^3$,which rearranges to $am^3 + (2a - h)m + k = 0$.This cubic equation in $m$ has three roots $m_1, m_2, m_3$,corresponding to the three normals drawn from the point $(h, k)$ to the parabola.The coordinates of the feet of these normals are $(am_i^2, -2am_i)$ for $i = 1, 2, 3$.The centroid $(x_c, y_c)$ of the triangle formed by these three points is given by:$x_c = \frac{a(m_1^2 + m_2^2 + m_3^2)}{3}$ and $y_c = \frac{-2a(m_1 + m_2 + m_3)}{3}$.From the cubic equation $am^3 + 0m^2 + (2a - h)m + k = 0$,we have the sum of roots $m_1 + m_2 + m_3 = 0$.Since $m_1 + m_2 + m_3 = 0$,it follows that $y_c = 0$.The condition $y_c = 0$ implies that the centroid lies on the axis of the parabola.

The centroid of the triangle formed by joining the feet of the normals drawn from any point to the parabola $y^2 = 4ax$ lies on

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