The normals at three points P, Q, and R of the parabola $y^{2} = 4 a x$ meet in a point O whose coordinates are h and k; prove that
If P be fixed, then QR is fixed in direction and the locus of the centre of the circle circumscribing PQR is a straight line.

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Solution

the slope of $Q R$
$= \frac{- 2 a [m_{2} - m_{3}]}{a [m_{2}^{2} - m_{3} 62]} = \frac{- 2}{m + 2 + m_{3}} = \frac{2}{m_{1}}$
If $m_{1}$ is constant $\frac{2}{m_{1}}$ is also constant, hence the direction $Q R$ is fixed.
The centre of the circle given
$(- g, - f)$ i.e., $(\frac{h + 2 a}{a}, \frac{k}{4})$ .
so if $(x, y)$ is the centre, $x = \frac{h + 2 a}{a}$ ...(1)
and $y = \frac{k}{4}$ ...(2)
again the equation of normal at $P \equiv (a m_{1}^{2} - 2 a m_{1})$ is
$y = m_{1} x - 2 a m_{1} - a m_{1}^{3}$ ...(3)
As this passes through $(h, k)$ and also $m_{1} =$ constant, then $k = m_{1} h - 2 a m_{1} - a m_{1}^{3}$ . Eliminating $h$ and $k$ from (1), (2) and (3), we have $4 y = m_{1} (2 x - 2 a) - 2 a m_{1} - a m_{1}^{3}$ . this is the locus of centre $(x m y)$ and it is clearly a straight line.

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