PR and QR are chords of a parabola which are normals at P and Q. Prove that two of the common chords of the parabola and the circle circumscribing the triangle PRQ meet on the directrix.

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Solution

Suppose the co-ordinates of $R$ on the parabola
$y^{2} = 4 a x$ be $(h, k)$ and let $h = a m_{2}^{2}$ and $k = - 2 a m_{3} .$
Again, let the co-ordinates of $P$ and $Q$ be $(a m_{1}^{2}, - 2 a m_{1})$ and $(a m_{2}^{2}, - 2 a m_{2})$ respectively. Now, if any normal $y = m x - 2 a m - a m^{3}$ passes through $(h, k)$ . Then we have
$m_{1} + m_{2} + m_{3} = 0, m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{h}$
and $m_{1} m_{2} m_{3} = \frac{- k}{a}$ .
But as $m_{3} = \frac{- k}{2 a}$ , so $m_{1} m_{2} = 2$ and $m_{1} + m_{2} = \frac{k}{2 a}$ . ...(1)
We have proved that the circle and parabola intersect in four points and that the line joining one pair of these four points and the line joining the other pair are equally inclined to the axis. Therefore , if S be the fourth point, then $R Q$ and $R S$ are equally inclined with the axis.
Again, the equation to $P Q$ will be $- y (m_{1} = m_{2}) = 2 x + 2 a m_{1} m_{2}$
$\Rightarrow - y \frac{k}{2 a} = 2 x + 4 a$ [from (1)]
$\to - k y = 4 a x + 8 a A^{2}$ . ...(2)
The equation to $R S$ is $k (y - k) = 4 a (x - h)$ or $y k = 4 a x + k^{2} - 4 a h$ .
But $k^{2} = 4 a h$ . So $y k = 4 a x$ . ...(3)
Adding (3) with (2), we have
$8 a x k + 8 a^{2} = 0$ or $x + a = 0$ .
This is the equation of directrix. Hence the common chords $P R$ and $R S$ of parabola an circle meet in directix.

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