Question

Find the locus of a point $O$ when the three normals drawn from it are such that two of them make angles with the axis the product of whose tangents is $2$.

Answer 1

The equation of normal to the parabola is given by $y=mx-2am-a{m}^3$

Three normals pass through the point $(h,k)$

Substituting $(h,k)$ into the equation, we have $a{m}^3+m(2a-h)+k=0$ with roots $m_1,m_2,m_3$

Since two normals make with the axis such that $m_1{m}_2=2$

Since $m_1{m}_2{m}_3=\frac{-k}{a},m_3=\frac{-k}{2a}$

Substituting $m_3$ into the equation, we get

$a\times \frac{-k^3}{8a^3}+(2a-h)\times \frac{-k}{2a}+k=0$

$\Rightarrow k^3+4ak(2a-h)-8k{a}^2=0$

i.e. $y^2+8a^2-4ax-8a^2=0$

i.e. $y^2-4ax=0$ is the required locus