Question

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

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 complementary angles with the axis, $m_1{m}_2=-1$

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

Substituting $m_3$ into the equation, we get

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

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

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

i.e. $y^2-ax+3a^2=0$ is the required locus