Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 48

Show that the family curve $y = ax^3 , x^2 + 3y^2 = b$ are orthogonal trajectories of each other, that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes.

Taking the derivative of the function $y = ax^3$ implicitly we have

$\displaystyle \frac{dy}{dx} = 3ax^2$

So the slope of the tangent line to this curve at point $(x_1, y_1)$ is


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& 3a (x_1)^2
\\
\\
\frac{dy}{dx} =& 3ax_1^2

\end{aligned}
\end{equation}
$


Similarly, on the other curve $x^2 + 3y^2 = b$


$
\begin{equation}
\begin{aligned}

2x + 3 (2y) \frac{dy}{dx} =& 0
\\
\\
\frac{dy}{dx} =& \frac{-x}{3y}


\end{aligned}
\end{equation}
$


So the slope at point $(x_1 , y_1)$ is

$\displaystyle \frac{dy}{dx} = \frac{-x_1}{3y_1} $

but $y_1 = ax_1^3$, so the slope of the second curve is

$\displaystyle \frac{dy}{dx} = \frac{-x_1}{3(ax_1^3)} = \frac{-1}{3ax_1^2} $

We know that if the two curves are orthogonal to each other, their tangent lines are perpendicular at each point of intersection, that is, the product of their slopes is equal to $-1$. Multiplying the slopes we get..

$\displaystyle (\cancel{3ax_1^2}) \left( \frac{-1}{\cancel{3ax_1^2}} \right) = -1$


Therefore, the curve must be orthagonal.