# Orthogonal Trajectories

Consider a one-parameter family of curves in the xy-plane defined by ${F}{\left({x},{y},{c}\right)}={0}$, where ${c}$ denotes the parameter. The problem is to find another one-parameter family of curves called the orthogonal trajectories of the family defined above and given analytically by ${G}{\left({x},{y},{k}\right)}={0}$, such that every curve in this new family intersects at right angles every curve in the original family.

We implicitly differentiate the first equation with respect to ${x}$, then eliminate ${c}$ between this derived equation and the equation ${F}{\left({x},{y},{c}\right)}={0}$. This gives an equation connecting ${x}$, ${y}$, and ${y}'$, which we solve for ${y}'$ to obtain a differential equation of the form $\frac{{{d}{y}}}{{{d}{x}}}={f{{\left({x},{y}\right)}}}$.

The orthogonal trajectories then are the solutions of $\frac{{{d}{y}}}{{{d}{x}}}=-\frac{{1}}{{{f{{\left({x},{y}\right)}}}}}$.

For many families of curves, one cannot explicitly solve for $\frac{{{d}{y}}}{{{d}{x}}}$ and obtain a differential equation of the form $\frac{{{d}{y}}}{{{d}{x}}}={f{{\left({x},{y}\right)}}}$.

Example 1. Find the orthogonal trajectories of the family of curves ${{x}}^{{2}}+{{y}}^{{2}}={{c}}^{{2}}$.

Rewrite as ${{x}}^{{2}}+{{y}}^{{2}}-{{c}}^{{2}}={0}$, or ${F}{\left({x},{y},{c}\right)}={{x}}^{{2}}+{{y}}^{{2}}-{{c}}^{{2}}={0}$. This family of curves consists of circles with centers at the origin and radii ${c}$. Implicitly differentiating the given equation with respect to ${x}$, we obtain that ${2}{x}+{2}{y}\frac{{{d}{y}}}{{{d}{x}}}={0}$, or $\frac{{{d}{y}}}{{{d}{x}}}=-\frac{{x}}{{y}}$.

Here, ${f{{\left({x},{y}\right)}}}=-\frac{{x}}{{y}}$; so, the orthogonal trajectories satisfy the equation $\frac{{{d}{y}}}{{{d}{x}}}=\frac{{y}}{{x}}$. This equation is separable: $\frac{{{d}{y}}}{{y}}=\frac{{{d}{x}}}{{x}}$, or ${\ln{{\left({y}\right)}}}={\ln{{\left({x}\right)}}}+{a}$, which can be rewritten as ${y}={k}{x}$, where ${k}={{e}}^{{a}}$.

The solution ${y}={k}{x}$ represents the orthogonal trajectories (straight lines that pass through the origin).

Let's solve another example.

Example 2. Find the orthogonal trajectories of the family of curves ${{y}}^{{2}}={4}{c}{x}$.

Rewrite as ${{y}}^{{2}}-{4}{c}{x}={0}$, or ${F}{\left({x},{y},{c}\right)}={{y}}^{{2}}-{4}{c}{x}={0}$. Implicitly differentiating the given equation with respect to ${x}$, we obtain that ${2}{y}\frac{{{d}{y}}}{{{d}{x}}}-{4}{c}={0}$, or $\frac{{{d}{y}}}{{{d}{x}}}=\frac{{{2}{c}}}{{y}}$.

Note that, from the initial equation, ${c}=\frac{{{{y}}^{{2}}}}{{{4}{x}}}$; so, $\frac{{{d}{y}}}{{{d}{x}}}=\frac{{{2}{\left(\frac{{{{y}}^{{2}}}}{{{4}{x}}}\right)}}}{{y}}$, or $\frac{{{d}{y}}}{{{d}{x}}}=\frac{{y}}{{{2}{x}}}$.

Here, ${f{{\left({x},{y}\right)}}}=\frac{{y}}{{{2}{x}}}$; so, the orthogonal trajectories satisfy the equation $\frac{{{d}{y}}}{{{d}{x}}}=-{2}\frac{{x}}{{y}}$. This equation is separable: it can be rewritten as ${y}{d}{y}=-{2}{x}{d}{x}$, or $\frac{{{{y}}^{{2}}}}{{2}}=-{{x}}^{{2}}+{k}$.

So, the orthogonal trajectories are ${{x}}^{{2}}+\frac{{{{y}}^{{2}}}}{{2}}={k}$ $\left({k}>{0}\right)$.