The normals at three points $P, Q, R$ of the parabola $y^{2} = 4 a x$ meet in $(h, k)$ . Prove that the centroid of triangle $P Q R$ on axis at distance $\frac{2}{3} (h - 2 a)$ from the vertex.

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Solution

a] Equation of normals to $y^{2} = 4 a x$ is
$y = m x - 2 a m - a m^{3}$
It passes through (h,k)
$∴ k = m h - 2 a m - a m^{3}$
$m_{1} + m_{2} + m_{3} = 0; m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - n}{a}$
Let P,Q,R be beet of there normal's
$P (a m^{2} - 2 a m) Q (a m_{2}^{2} - 2 a m_{2}) R (a m_{3}^{2} - 2 a m_{3})$
$¯ x = \frac{a}{3} (m_{1}^{2} + m_{2}^{2} + m_{3}^{2}) = \frac{a}{3} [a^{2} - 2 (\frac{20 - h}{a})]$
$= \frac{2}{3} (h - 2 a)$
$¯ y = \frac{- 2 a}{3} (m_{1} + m_{2} + m_{3}) = 0$ Hence, proved

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