Find the locus of a point $O$ when the three normals drawn from it are such that the area of the triangle formed by their feet is constant.

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Solution

Suppose the co-ordinates of the feet of the normals $A, B$ and $C$ be
$(a m_{1}^{2}, - 2 a m_{2}), (a m_{2}^{2}, - 2 a m_{2})$ and $(a m_{3}^{2}, - 2 a m_{3})$ respectively.
Hence, the area of the triangle $A B C$ is given as
$= \frac{1}{2} [a m_{1}^{2} (- 2 a m_{2} + 2 a m_{3}) + a m_{2}^{2} (- 2 a m_{3} + 2 a m_{1}) + a m_{2}^{2} (- 2 a m_{1} + 2 a m_{2})]$
$= - a^{2} [m_{1}^{2} (m_{2} - m_{3}) + m_{2}^{2} (m_{3} - m_{1}) + m_{3}^{2} (m_{1} - m_{2})]$
$\Rightarrow a^{2} (m_{1} - m_{2}) (m_{2} - m_{3}) (m_{3} - m_{1}) =$ constant (by hypothesis)
$∴ (m_{1} - m_{2})^{2} (m_{2} - m_{3})^{2} (m_{3} - m_{1})^{2} =$ constant $= λ$ (say) ...(1)
But $(m_{1} - m_{2})^{2} = (m_{1} + m_{2})^{2} - 4 m_{1} m_{2} = (- m_{3})^{2} + \frac{4 k}{a m_{3}}$
$= \frac{1}{a m_{3}} (a m_{3}^{3} + 4 k)$

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