Question

Find the locus of the point N from which 3 normals are drawn to the parabola $y^2=4ax$ are such that two of them are perpendicular to each other

Answer 1

Equation of normal to $y^2=4ax$ is

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

Let the normal passes through N(h, k)

$\therefore k=mh-2am-a{m}^3\Rightarrow a{m}^3+(2a-h)m+k=0$

For given value's of (h, k) it is cubic in 'm' Let $m_1,m_2\&m_3$ are root's of above equation

$\therefore m_1+m_2+m_3=\mathrm{0........}\left(i\right)$

$m_1{m}_2+m_2{m}_3+m_3{m}_1=\frac{2a-h}{a}.........\left(ii\right)$

$m_1{m}_2{m}_3=-\frac{k}{a}.........\left(iii\right)$

If two normal's are perpendicular

$\therefore m_1{m}_2=-1From\left(3\right)m_3=\frac{k}{a}.........\left(iv\right)$

$From\left(2\right)-1+\frac{k}{a}(m_1+m_2)=\frac{2a-h}{a}........\left(v\right)$

$From\left(1\right)m_1+m_2=-\frac{k}{a}........\left(vi\right)$

from (5) & (6) we get

$-1-\frac{k^2}{a}=2-\frac{h}{a}\Rightarrow y^2=a(x-3a)$