# 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 $$${{x}}^{{2}}+\frac{{{{y}}^{{2}}}}{{2}}={k}$$$ $$$\left({k}>{0}\right)$$$.