Careers360 Logo
Is JEE Main syllabus reduced for 2025

Orthogonal Trajectory - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 16 Questions around this concept.

Concepts Covered - 1

Orthogonal Trajectory

Suppose that a family of curves f (x, y, a) = 0 is given. A curve that intersects every member of this family of curves at 90o, then it is called an orthogonal trajectory of the given family of curves.

Consider the family of all circles having their center at origin (a few such circles appear in figure below)

 

The orthogonal trajectories for this family of circles would be members of the family of straight lines passing through the origin (shown by dashed lines).

 

Steps for finding Orthogonal Trajectory :

  1. The equation of the family of curves is f(x, y, a) = 0, where ‘a’ is an arbitrary constant. Differentiate ‘f’  with respect to ‘x’ and eliminate ‘a’.

  2. The differential equation you get in step 1, substitute, \\\mathrm{-\frac{dx}{dy}\;\;for\;\;\frac{dy}{dx}}. This will give the differential equation of the orthogonal trajectory.

  3. By solving this differential equation, we get the required orthogonal trajectory.


Illustration:

The orthogonal trajectories of the family y = x + ce-x

Differentiation of the given equation gives

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\frac{dy}{dx}=1-ce^{-x}}\\\\\text{Eliminate the arbitrary constant 'c'}\\\\\Rightarrow \;\;\;\;\;\;\;\mathrm{\frac{dy}{dx}=1+x-y}\\\text{Thus, the differential equation for the family of orthogonal trajectories is }\\\\\mathrm{\;\;\;\;\;\;\;\;\;-\frac{dx}{dy}=1+x-y}\\\\\Rightarrow \;\;\;\;\;\;\; \frac{d x}{d y}+x=y-1\\\\\text{which is a linear differential equation having the solution as}\\\mathrm{\;\;\;\;x e^{y}-e^{y}(y-2)=c_{1}}

Study it with Videos

Orthogonal Trajectory

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Books

Reference Books

Orthogonal Trajectory

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 10.17

Line : 11

E-books & Sample Papers

Get Answer to all your questions

Back to top