**Trajectory**

A curve which cuts every member of a given family of curves in accordance with some given law, is called trajectory of the given family of curves.

**Orthogonal Trajectory**

If a cuts every member of given family of curves at right angle it is called orthogonal trajectory.

As an example y = mx and 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

**Determination of orthogonal trajectories in cartesiancoordinates**

**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 be the angle, which the tangent at P to the member PQ with x axis.

Then …(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 , let be the angle which the tangent to the trajectory makes with x axis.

So, …(4)

Let PT and PT’ intersect at 90°.

So

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

So , is the required family of trajectories.

**Determination of orthogonal trajectories in polarcoordinate system**

Let the equation of given family of curves be in polar form …(1)

Where c is a parameter

Differentiate (1) w.r.t and eliminating c between (1) and derived equation. We get the differential

equation of the given family of curves.

Let be that differential equation …(2)

Let be the angle between the tangent PT to a member PQ of the given family of curves and radius vector OP at any point .

then …(3)

Let 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 be the angle which the tangent PT’ to the trajectory makes with the common radius vector OP.

Then

Let PT and PT’ intersect at 90°

Then

So using (3) and (4)

At the point of intersection r = r’ and

So is the required orthogonal trajectory.

**Determination of Orthogonal trajectories in PolarCoordinates**

**Working Rule :**

1. Differentiate the given equation of family of curves w.r.t (generally take logarithm). Eliminate the parameter.

2. Replace by and obtain the differential equation of orthogonal trajectories

3. Obtain the general solution of differential equation obtained above.

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