Orthogonal Trajectory - Practice Questions & MCQ

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 $90^{\circ}$, 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 the 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 :

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’.

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

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
$
\frac{\mathrm{dy}}{\mathrm{dx}}=1-\mathrm{ce}^{-\mathrm{x}}
$

Eliminate the arbitrary constant 'c'
$
\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=1+\mathrm{x}-\mathrm{y}
$

Thus, the differential equation for the family of orthogonal trajectories is
$
\begin{aligned}
& -\frac{\mathrm{dx}}{\mathrm{dy}}=1+\mathrm{x}-\mathrm{y} \\
\Rightarrow \quad & \frac{d x}{d y}+x=y-1
\end{aligned}
$
which is a linear differential equation having the solution as
$
\mathrm{xe}^{\mathrm{y}}-\mathrm{e}^{\mathrm{y}}(\mathrm{y}-2)=\mathrm{c}_1
$