What are Orthogonal Trajectories?

Example

Let's say we are to find the orthogonal trajectory of the family of equilateral hyperbolas: \(xy=C\)

We divide our solution into three parts.

First, we express the original family in terms of its first derivative:

• \(x \frac{d y}{d x}+y=0\)
• \(\frac{d y}{d x}=\frac{-y}{x}=y^{\prime}\)

Second, we then find the slope of the orthogonal curve by using the relationship:

• \(y^{\prime} y_{\perp}^{\prime}=-1\)

From the first part, we know that \(y^{\prime}=\frac{-y}{x}\). All we have to do is solve for \(y_{\perp}^{\prime}\).

• \(\frac{-y}{x} y_{\perp}^{\prime}=-1\)
• \(y_{\perp}^{\prime}=\frac{x}{y}\)

Third, we find the equation of the orthogonal curves by solving for the differential equation.

• \(\frac{d y}{d x}=\frac{x}{y}\)

Apply separation of variables:

• \(\int y d y=\int x d x\)
• \(\frac{1}{2} y^2=\frac{1}{2} x^2+C_1\)
• \(y^2=x^2+2 C_1\), let \(2 C_1=K\)
• \(y^2=x^2+K\)

The orthogonal family to the equilateral hyperbolas, \(xy=C\), is \(y^2=x^2+K\). Below is a plot of the two families.

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The blue curves represent the family of equilateral hyperbolas, while the red curves represent its orthogonal trajectory. At each intersection, the curves meet at right angles.

Summary

When two curve families are perpendicular to each other at meeting points, we say that these are mutually orthogonal curves.

We say that two families are orthogonal to each other if the product of their first derivatives is -1. From this, we can find the orthogonal trajectory of any given curve.