Orthogonal Trajectory: Definition, Questions and Answers

An orthogonal trajectory is a curve that intersects another family of curves at right angles. In this post, we study orthogonal trajectory along with a few questions and answers.

Definition of Orthogonal Trajectory

An orthogonal trajectory of a family of curves is a curve that intersects every member of that family at a right angle.

For example, the straight line y=x is an orthogonal trajectory of the family of circles x²+y²=a² (see the example below).

How to Find Orthogonal Trajectory

Method of Finding Orthogonal Trajectories: We know that if a line intersects another line then the product of their slopes is -1. As an orthogonal trajectory intersects a family of curves at right angles, we have to replace the slope dy/dx according to the form of the curve as follows:

• For cartesian form: replace $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$

• For polar form: replace $\dfrac{dr}{d\theta}$ by $-r^2\dfrac{d\theta}{dr}$

• For ω-trajectories, replace $\dfrac{dy}{dx}$ by $\dfrac{\frac{dy}{dx} -\tan \omega}{1+\frac{dy}{dx} \tan \omega}$.

Orthogonal Trajectory Questions and Answers

Using the above method, let us learn how to compute orthogonal trajectory of various curves, like straight lines, circles, parabolas, etc.

Orthogonal Trajectory of Straight Lines

$\boxed{\color{blue}\textbf{Question 1}:}$ Find the orthogonal trajectories of the straight lines y=mx, where m is a parameter.

$\boxed{\color{red}\textbf{Answer:}}$

Differentiating both sides of y=mx with respect to x, we have:

$\dfrac{dy}{dx} =m = \dfrac{y}{x}$ as y=mx.

Replacing $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$, we get that

$-\dfrac{dx}{dy}=\dfrac{y}{x}$

⇒ xdx + ydy =0

Integrating, we have x²+y²=c², where c is a parameter.

This family x²+y²=c² of circles passing through origin is the orthogonal trajectories of the family of straight lines y=mx.

Orthogonal Trajectory of Hyperbola

$\boxed{\color{blue}\textbf{Question 2}:}$ Find the orthogonal trajectories of the family of hyperbolas xy=a² where a is a parameter.

$\boxed{\color{red}\textbf{Answer:}}$

Differentiating both sides of xy=a² with respect to x, we have:

$y+x\dfrac{dy}{dx}=0$

⇒ $\dfrac{dy}{dx}=-\dfrac{y}{x}$

Now, replacing $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$, we get that

$-\dfrac{dx}{dy}=-\dfrac{y}{x}$

⇒ $\dfrac{dx}{dy}=\dfrac{y}{x}$

⇒ xdx – ydy =0

Integrating, x²-y²=c².

Therefore, the orthogonal trajectories of the family of curves xy=a² are given by the family of curves x²-y²=c².

Orthogonal Trajectory of Parabolas

$\boxed{\color{blue}\textbf{Question 3}:}$ Find the orthogonal trajectories of the family of parabolas y²=2ax, where a is a parameter.

$\boxed{\color{red}\textbf{Answer:}}$

Differentiating both sides of y²=2ax w.r.t x, we get that

$2y\dfrac{dy}{dx} =2a$

⇒ $y\dfrac{dy}{dx} =a=\dfrac{y^2}{2x}$, since y²=2ax.

⇒ $\dfrac{dy}{dx} =\dfrac{y}{2x}$

Now, replace $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$, so we obtain the following:

$-\dfrac{dx}{dy}=\dfrac{y}{2x}$

⇒ ydy + 2xdx =0

Integrating both sides, we get x²+y²/2=k.

Lets rewrite the family as follows 2x²+y²=c², where k=c².

This family 2x²+y²=c² of concentric ellipses is the orthogonal trajectories of the family of parabolas y²=2ax.

Orthogonal Trajectory of x^2/3+y^2/3=a^2/3

$\boxed{\color{blue}\textbf{Question 4}:}$ Find the orthogonal trajectories of the family of curves x^2/3+y^2/3=a^2/3, where a is a parameter.

$\boxed{\color{red}\textbf{Answer:}}$

Differentiating both sides x^2/3+y^2/3=a^2/3,

2/3 x^-1/3 + 2/3 y^-1/3=0

⇒ $\dfrac{dy}{dx} = -\left(\dfrac{y}{x}\right)^{1/3}$

Now, replacing $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$, it follows that

$-\dfrac{dx}{dy} = -\left(\dfrac{y}{x}\right)^{1/3}$

⇒ $\dfrac{dx}{dy} = \left(\dfrac{y}{x}\right)^{1/3}$

This implies that

x^1/3dx = y^1/3dy

Integrate both sides. So we have:

3/4 x^4/3 – 3/4 y^4/3 = 3/4 c^4/3

⇒ x^4/3 – y^4/3 = c^4/3. This family of curves is the orthogonal trajectories of the family of curves x^2/3+y^2/3=a^2/3.