**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=a^{2}.

**Answer:**

xy=a^{2}

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, x^{2}-y^{2}=c^{2}.

This is the orthogonal trajectories of the family of curves xy=a^{2}.

**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, x^{2}+y^{2}=c^{2}.

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

**Question3:** Find the orthogonal trajectories of the curves y^{2}=4ax.

**Answer:**

y^{2}=4ax

Differentiating both sides w.r.t x,

2y$\frac{dy}{dx}$ =4a = y^{2}/4a

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

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

⇒ y^{2}dy + 2xydx =0

⇒ ydy + 2xdx =0

Integrating, x^{2}+y^{2}/2=c. This is the desired orthogonal trajectories of y^{2}=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^{1/3}dx = y^{1/3}dy

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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**Abel’s Theorem of Series Convergence**

**Rolle’s Theorem Questions-Answers**