If a cuts every member of given family of curves at right angle it is called orthogonal trajectory.
As an example y = mx and $x^2 + y^2 = a^2$ are respectively the family of straight line and family of circles with centre or origin and radius a.
Every line y = mx cut each member of family of circles at right angle.

So y = mx is orthogonal trajectory of $x^2 + y^2 = a^2$

Determination of orthogonal trajectories in cartesian
coordinates

Working Rule.
1. Differentiate the given equation of family of curves. Eliminate the parameters between the derived equation and given equation of the family. It will give the differential equation of given family of curve.
2. Replace dy/dx (slope of tangent) by –dx/dy (slope of orthogonal family) in the above differential equation in step 1.
3. Obtain the general solution of differential equation in step 2.

Self orthogonal family of Curves

If each member of a given family of curves intersects all other members orthogonally, then given family of curves is said to be self orthogonal.
Let the equation of given family of curves be f(x, y, c) = 0 …(1)

Differentiate (1) w.r.t x and eliminate c between (1) and derived equation
We shall arrive at the differential equation of given family as F(x, y, dy/dx) = 0 …(2)

Let $\psi$ be the angle, which the tangent at P to the member PQ with x axis.

Then $\tan \psi = \dfrac{{dy}}{{dx}}$ …(3)

Let (X, Y) be the current coordinates of any point of a trajectory. At point of intersection P of any members of (2) with the trajectory $PQ'$ , let $\psi'$ be the angle which the tangent $PT'$ to the trajectory makes with x axis.

So, $\tan \psi ' = \dfrac{{dY}}{{dX}}$ …(4)

Let PT and PT’ intersect at 90°.

So $\tan \psi \times \tan \psi' = -1 \Rightarrow \dfrac{{dy}}{{dx}} \cdot \dfrac{{dY}}{{dX}} = -1$

$\Rightarrow \ \ \ \ \ \ \dfrac{{dY}}{{dX}} = - \dfrac{{dx}}{{dy}}$

At the point of intersection x = X and y = Y

So $F \; \left( {X,\;Y,-\; \dfrac{{dx}}{{dy}}} \right) = 0$ , is the required family of trajectories.

Determination of orthogonal trajectories in polar
coordinate system

Let the equation of given family of curves be $f(r,\;\theta ,\;c)\; = \;0$ in polar form …(1)

Where c is a parameter

Differentiate (1) w.r.t $\theta$ and eliminating c between (1) and derived equation. We get the differential
equation of the given family of curves.

Let $F(r,\theta, dr/ d \theta) = 0$ be that differential equation …(2)

Let $\phi$ be the angle between the tangent PT to a member PQ of the given family of curves and radius vector OP at any point $P(r,\;\theta )$ .

then $\tan \phi = r\left( {\dfrac{{d\theta }}{{dr}}} \right)$ …(3)

Let $(r', \theta')$ be the current coordinates of any point of a trajectory. At point of intersection P of any member of (2) with the trajectory PQ’, let $\phi'$ be the angle which the tangent PT’ to the trajectory makes with the common radius vector OP.

Then $\tan \phi ' = r'{(d\theta ' / dr')}$

Let PT and PT’ intersect at 90°

Then $\phi '-\phi = 90 \Rightarrow \phi ' = 90 + \phi$

$\tan \phi' = \tan (90 + \phi) = -\cot \phi \Rightarrow \tan \phi',\tan \phi = -1$

So using (3) and (4)

$\left({r \dfrac{{d\theta }}{{dr}}} \right)\left( {r'\dfrac{{d\phi '}}{{dr'}}} \right) = -1$

$\dfrac{{dr}}{{d\theta }} = -r \cdot r' \dfrac{{d\theta '}}{{dr'}}$

At the point of intersection r = r’ and $\theta = \theta'$

So $F\left( {r,\theta ,- {r^2}\frac{{d\theta }}{{dr}}} \right) = 0$ is the required orthogonal trajectory.

Determination of Orthogonal trajectories in Polar
Coordinates $f{\rm{(}}r,{\theta_{\rm{1}}}c{\rm{)}} = {\rm{0}}$

Working Rule :
1. Differentiate the given equation of family of curves w.r.t $\theta$ (generally take logarithm). Eliminate the parameter.
2. Replace $(dr/d \theta)$ by $-r^2 (d\theta / dr)$ and obtain the differential equation of orthogonal trajectories
3. Obtain the general solution of differential equation obtained above.

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