Find the locus of a point $O$ when the three normals drawn from it are such that two of them make equal angles with the given line $y = m x + c$ .

Open in App

Solution

Three normals can be drawn whose slopes are$ {m}_{1},{m}_{2} and {m}_{3}
For cubic $\Rightarrow m_{1} + m_{2} + m_{3} = 0 m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = (\frac{2 a - h}{a}) m_{1} m_{2} m_{3} = \frac{- k}{a}$
The equation o normal $y^{2} = 4 a x i . e, y = m x - 2 a m - a m^{3}$
It passesthrough the point (h,t) if
$K = m h - 2 a m - a m^{3} a m^{2} + m (2 a - h) + k = 0 ⟶ (1)$
Let the roots be $m_{1} {, m}_{2} a n d m_{3}$ .The perpendicular normals.Correspond to the values of $m_{1} {, m}_{2}$
from the eqn, $m_{1} m_{2} m_{3} = \frac{- k}{a}$
Since $m_{3}$
$⟹ {(\frac{K}{a})}^{3} + \frac{K}{a} (2 a - h) + K = 0 ⟹ K^{2} + a (2 a - h) + a^{2} = 0 ∴ K^{2} = a (h - 3 a)$
$∴$ Locus of $(h, k)$ is $y^{2} = a (x - 3 a)$

Suggest Corrections

Similar questions

O

O

2

P

y^{2} = 4 a x

O

O

Join BYJU'S Learning Program