Question

Find the locus of the point of intersection of two normals to a parabola which are at right angles to one another.

Answer 1

The equation of the normal to the parabola $y^2=4ax$ is $y=mx-2am-a{m}^3$.

It passes through the point $(h,k)$ is

$k=mh-2am-a{m}^3$

$a{m}^3+m(2a-h)+k=0$ .....(1)

Let the roots of the above equation be $m_1,m_2,m_3$. Let the perpendicular normals correspond to the values of $m_1$ and $m_2$ so that $m_1{m}_2=-1$.

From equation 1, $m_1{m}_2{m}_3=-\frac{k}{a}$

Since, $m_1{m}_2=-1,m_3=\frac{k}{a}$

Since $m_3$ is a root of equation 1, we have

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

$k^2+a(2a-h)+a^2=0$

$k^2=a(h-3a)$

Hence, the locus of $(h,k)$ is $y^2=a(x-3a)$.