Orthogonal Trajectory with Solved Problems

Definition of orthogonal trajectory

Let us consider the two families of curves with the property: every member of either family cuts each member of the other family at right angles. Such two families of curves are called orthogonal trajectories of each other.

Method of Finding orthogonal trajectories

• For cartesian form: replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$
• For polar form: replace $\frac{dr}{d\theta}$ by $-r^2\frac{d\theta}{dr}$
• For ω-trajectories, replace $\frac{dy}{dx}$ by $\dfrac{\frac{dy}{dx} -\tan \omega}{1+\frac{dy}{dx} \tan \omega}$.

Question-Answer on orthogonal trajectory

Question1: Find the orthogonal trajectories of the curves xy=a².

Answer:

xy=a²

Differentiating both sides w.r.t x,

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

Now, replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$. So we get that

y – x$\frac{dx}{dy}$ =0

⇒ xdx – ydy =0

Integrating, x²-y²=c².

This is the orthogonal trajectories of the family of curves xy=a².

Question2: Find the orthogonal trajectories of the straight lines y=mx.

Answer:

y=mx

Differentiating both sides w.r.t x,

$\frac{dy}{dx}$ =m = y/x

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

$-\frac{dx}{dy}$ =y/x

⇒ xdx + ydy =0

Integrating, x²+y²=c².

This is the orthogonal trajectories of the family of straight lines y=mx.

Question3: Find the orthogonal trajectories of the curves y²=4ax.

Answer:

y²=4ax

Differentiating both sides w.r.t x,

2y$\frac{dy}{dx}$ =4a = y²/4a

Replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$. Thus,

$-2y\frac{dx}{dy}=\frac{y^2}{x}$

⇒ y²dy + 2xydx =0

⇒ ydy + 2xdx =0

Integrating, x²+y²/2=c. This is the desired orthogonal trajectories of y²=4ax that defines a family of concentric ellipses.

Question4: Find the orthogonal trajectories of the curves x^2/3+y^2/3=a^2/3, where a is a parameter.

Answer:

x^2/3+y^2/3=a^2/3

Differentiating both sides w.r.t x,

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

⇒ dy/dx = -(y/x)^1/3

Replacing $\frac{dy}{dx}$ by $-\frac{dx}{dy}$, we obtain that

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

⇒ x^1/3dx = y^1/3dy

Integrating, 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 is the desired orthogonal trajectories of x^2/3+y^2/3=a^2/3.

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