If the normals at the three points P, Q, and R meet in a point and if PP', QQ', and RR' be chords parallel to QR, RP, and PQ respectively, prove that the normals at P', Q', and R' also meet in a point.

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Solution

Let the feet of normals at $P, Q$ and $R$ be $(a m_{1}^{2}, - 2 a m_{1}), (a m_{2}^{2}, - 2 a m_{2})$ and $(a m_{3}^{2}, - 2 a m_{3})$ respectively.

And the feet of normals at $P^{'}, Q^{'}$ and $R^{'}$ be $(a t_{1}^{2}, - 2 a t_{1}), (a t_{2}^{2}, - 2 a t_{2})$ and $(a t_{3}^{2}, - 2 a t_{3})$ respectively.

Slope of $P P^{'} = \frac{- 2 a m_{1} + 2 a t_{1}}{a m_{1}^{2} - a m_{2}^{2}} = \frac{- 2}{m_{1} + t_{1}}$

Slope of $Q R = \frac{- 2 a m_{2} + 2 a m_{3}}{a m_{2}^{2} - a m_{3}^{2}} = \frac{- 2}{m_{2} + m_{3}}$

$P P^{'}$ is parallel to $Q R$

$∴ \frac{- 2}{m_{1} + t_{1}} = \frac{- 2}{m_{2} + m_{3}} m_{1} + t_{1} = m_{2} + m_{3}$

As $\sum m_{1} = 0$ ...... [Since, normals at $P, Q, R$ meet in a point]

$m_{1} + t_{1} = - m_{1} t_{1} = - 2 m_{1} . . . . . . . (i)$

Similarly,

$t_{2} = - 2 m_{2} . . . . . . . (i i) t_{3} = - 2 m_{3} . . . . . . . (i i i)$

Adding $(i), (i i)$ and $(i i i)$

$t_{1} + t_{2} + t_{3} = - 2 (m_{1} + m_{2} + m_{3}) \sum m_{1} = 0 t_{1} + t_{2} + t_{3} = 0 \Rightarrow \sum t_{1} = 0$

Hence proved that normals at $P^{'}, Q^{'}$ and $R^{'}$ also meet in a point.

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