Question

Find the locus of a point $O$ when the three normals drawn from it are such that the sum of the three angles made by them with the axis is constant.

Answer 1

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

Equation of normal in slope form is

$y=mx-2am-a{m}^3$

It passes through $O(h,k)$

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

This is cubic in $m$

$\therefore m_1+m_2+m_3=-\frac{0}{a}=0m_1{m}_2+m_2{m}_3+m_3{m}_1=\frac{2a-h}{a}{m}_1{m}_2{m}_3=-\frac{k}{a}$

Let ${\theta }_1{,\theta }_2,{\theta }_3$ be the angles made by them with the axis of the parabola

Given ${\theta }_1+{\theta }_2+{\theta }_3=constant$

$\tan {(\theta }_1{+\theta }_2,{+\theta }_3)=\tan \left(constamt\right)\frac{\tan {\theta }_1+\tan {\theta }_2+\tan {\theta }_3-\tan {\theta }_1\tan {\theta }_2\tan {\theta }_3}{1-\tan {\theta }_1\tan {\theta }_2-\tan {\theta }_2\tan {\theta }_3-\tan {\theta }_3\tan {\theta }_1}=c\frac{\tan {\theta }_1+\tan {\theta }_2+\tan {\theta }_3-\tan {\theta }_1\tan {\theta }_2\tan {\theta }_3}{1-(\tan {\theta }_1\tan {\theta }_2+\tan {\theta }_2\tan {\theta }_3+\tan {\theta }_3\tan {\theta }_1)}=c\frac{0-(-\frac{k}{a})}{1-\frac{2a-h}{a}}=c\frac{k}{h-a}=ck=hc-ac$

Reeplacing $h$ by $x$ and $k$ by $y$

$y=cx-ac$ where $c$ is a constant

represents a straight line.