Question

Find the orthogonal trajectories of the family of parabolas y² = 4ax, where a is a parameter.

Answer 1

The given curve is

y² = 4ax ...(i)

2y(dy/dx) = 4a

a = ydy/2dx ...(ii)

Eliminating a from Eqs (i) and (ii), we get 

y² = 4x(ydy/2dx)

y = 2x(dy/dx)

Replacing dy/dx by – dx/dy, we get

y = (2x) x -(dy/dx)

ydy + 2xdx = 0

Integrating, we get 

(y²/2) + x² = c

which is the required orthogonal trajectory.