Find the locus of a point $O$ when the three normals drawn from it are such that one bisects the angle between the other two.

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Solution

The equation of normal to the parabola is given by $y = m x - 2 a m - a m^{3}$
Three normals pass through the point $(h, k)$
Substituting $(h, k)$ into the equation, we have $a m^{3} + m (2 a - h) + k = 0$ with roots $m_{1}, m_{2}, m_{3}$
Since one normal bisects the angle between the other two, we can write $\frac{m_{3} - m_{2}}{1 + m_{3} m_{2}} = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$
$\Rightarrow m_{3} - m_{2} + m_{1} m_{2} m_{3} - m_{1} m_{2}^{2} = m_{2} - m_{1} + m_{3} m_{2}^{2} - m_{1} m_{2} m_{3}$
$\Rightarrow m_{3} + m_{1} - 2 m_{2} + 2 m_{1} m_{2} m_{3} - m_{2}^{2} (m_{1} + m_{3}) = 0$
Also, $m_{1} m_{2} m_{3} = \frac{- k}{a}$ and $m_{1} + m_{2} = - m_{3}$
$\Rightarrow - 3 m_{2} + \frac{- 2 k}{a} + m_{2}^{3} = 0$
Substituting $m_{2}^{3}$ into the equation, we get
$3 a m_{2} + 2 k + m_{2} (2 a - h) + k = 0$
$\Rightarrow 3 k + (5 a - h) m_{2} = 0$
i.e. $m_{2} = \frac{3 k}{h - 5 a}$
i.e. ${(\frac{3 k}{h - 5 a})}^{3} - \frac{9 k}{h - 5 a} - \frac{2 k}{a} = 0$
$\Rightarrow 27 a k^{2} - 9 a (h - 5 a)^{2} - 2 (h - 5 a)^{3} = 0$
$\Rightarrow 27 a k^{2} - 9 a h^{2} - 225 a^{3} + 90 a^{2} h - 2 h^{3} + 30 a h^{2} - 150 h a^{2} + 250 a^{3} = 0$
$\Rightarrow 27 a y^{2} + 21 a x^{2} - 60 a^{2} x - 2 x^{3} + 25 a^{3} = 0$

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