Find the locus of a point $O$ when the three normals drawn from it are such that the line joining the feet of two of them is always in a given direction.

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Solution

Let the point $O$ be $(h, k)$

Equation of normal in slope form is

$y = m x - 2 a m - a m^{3}$

It passes through $O (h, k)$

$a m^{3} + (2 a - h) m + k = 0$

This is cubic in $m$

$∴ m_{1} + m_{2} + m_{3} = - \frac{0}{a} = 0........ (i) m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a} . . . . . . . . . (i i) m_{1} m_{2} m_{3} = - \frac{k}{a} . . . . . . . . . . . (i i i)$

The feet of normal are $(a m_{1}^{2}, - 2 a m_{1}), (a m_{2}^{2}, - 2 a m_{2}), (a m_{3}^{2}, - 2 a m_{3})$

Given line joining any two is in a given direction

$\frac{- 2 a m_{2} - (- 2 a m_{1})}{a m_{2}^{2} - a m_{1}^{2}} = c \frac{2 a (m_{1} - m_{2})}{a (m_{2} + m_{1}) (m_{2} - m_{1})} = c m_{2} + m_{1} = - \frac{2}{c} m_{1} + m_{2} + m_{3} = 0 - \frac{2}{c} + m_{3} = 0 \Rightarrow m_{3} = \frac{2}{c} m_{1} m_{2} m_{3} = - \frac{k}{a} m_{1} m_{2} \frac{2}{c} = - \frac{k}{a} \Rightarrow m_{1} m_{2} = - \frac{k c}{2 a} m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a} - \frac{k c}{2 a} + m_{3} (m_{2} + m_{1}) = \frac{2 a - h}{a} - \frac{k c}{2 a} + \frac{2}{c} (- \frac{2}{c}) = \frac{2 a - h}{a} - \frac{k c}{2 a} - \frac{4}{c^{2}} = \frac{2 a - h}{a} \frac{- k c^{3} - 8 a}{2 a c^{2}} = \frac{2 a - h}{a} - k c^{3} - 8 a = 4 a c^{2} - 2 c^{2} h 2 c^{2} h - k c^{3} = 4 a c^{2} + 8 a$

Replacing $h$ by $x$ and $k$ by $y$

$2 c^{2} x - y c^{3} = 4 a c^{2} + 8 a$ $,$ where $c$ is a constant

This represents a straight line.

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