Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 46

Show that the family curve $x^2 + y^2 = ax, x^2 + y^2 = by$ 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 $x^2 + y^2 = ax$ implicitly we have

$\displaystyle 2x + 2y \frac{dy}{dx} = a$

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

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

$\displaystyle \frac{dy}{dx} = \frac{a - 2x_1}{2y_1}$

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


$
\begin{equation}
\begin{aligned}

2x + 2y \frac{dy}{dx} =& b \frac{dy}{dx}
\\
\\
\frac{dy}{dx} =& \frac{2x}{b - 2y}
\\
\\


\end{aligned}
\end{equation}
$


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

$\displaystyle \frac{dy}{dx} = \frac{2x_1}{b - 2y_1} $

but from the given equation $x_1^2 + y_1^2 = ax_1$ and $x_1^2 + y_1^2 = by_1$

$ax_1 = by_1$

$\displaystyle x_1 = \frac{by_1}{a}$ and $\displaystyle y_1 = \frac{ax-1}{b}$


Substituting these values to the given equation, we have


$
\begin{equation}
\begin{aligned}

& x_1^2 + y_1^2 = ax_1
\\
\\
& \left( \frac{by_1}{a} \right) ^2 + y_1^2 = \cancel{a} \left( \frac{by_1}{\cancel{a}} \right)
\\
\\
& \frac{b^2}{a^2} y_1^2 + y_1^2 = by_1
\\
\\
& y_1^{\cancel{2}} \left( \frac{b^2}{a^2} + 1 \right) = b \cancel{y_1}
\\
\\
& y_1 = \frac{b}{\left( 1 + \frac{b^2}{a^2} \right) }

\end{aligned}
\end{equation}
$


Consequently,


$
\begin{equation}
\begin{aligned}

x_1^2 + y_1^2 =& ax_1
\\
\\
x_1^2 + \left( \frac{ax_1}{b} \right)^2 =& ax_1
\\
\\
x_1^2 + \frac{a^2}{b^2} x_1^2 =& ax_1
\\
\\
x_1^{\cancel{2}} \left( 1 + \frac{a^2}{b^2} \right) =& a \cancel{x_1}
\\
\\
x_1 =& \frac{a}{\displaystyle \left( 1 + \frac{a^2}{b^2} \right)}

\end{aligned}
\end{equation}
$


Substituting $x_1$ and $y_1$ to the slope of the tangent lines we have @ $\displaystyle \frac{dy}{dx} = \frac{a - 2x_1}{2y_1}$,


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{\displaystyle a - 2 \left[ \frac{a}{\displaystyle 1 + \frac{a^2}{b^2}} \right] }{2 \left[ \displaystyle \frac{b}{\displaystyle 1 + \frac{b^2}{a^2}} \right] } = \frac{\displaystyle a - \frac{2a}{\displaystyle \frac{b^2 + a^2}{b^2}}}{\displaystyle \frac{2b}{\displaystyle \frac{a^2 + b^2}{a^2}}}
\\
\\
\\
\\
\frac{dy}{dx} =& \frac{\displaystyle a - \frac{2a (b^2)}{b^2 + a^2}}{\displaystyle \frac{2b(a^2)}{a^2 + b^2}} = \frac{\displaystyle \frac{a(b^2 + a^2) - 2a (b^2)}{\cancel{b^2 + a^2}}}{\displaystyle \frac{2b(a^2)}{\cancel{a^2 + b^2}}}
\\
\\
\\
\\
\frac{dy}{dx} =& \frac{ab^2 + a^3 - 2ab^2}{2ba^2} = \frac{\cancel{a} (b^2 + a^2 - 2b^2)}{2 \cancel{a^2} b}
\\
\\
\\
\\
\frac{dy}{dx} =& \frac{b^2 + a^2 - 2b^2}{2ab}
\\
\\
\frac{dy}{dx} =& \frac{a^2 - b^2}{2ab}


\end{aligned}
\end{equation}
$


@ $\displaystyle \frac{dy}{dx} = \frac{2x_1}{b - 2y_1} $



$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{\displaystyle 2 \left[ \frac{a}{\displaystyle 1 + \frac{a^2}{b^2}} \right]}{\displaystyle b - 2 \left[ \displaystyle \frac{b}{\displaystyle 1 + \frac{b^2}{a^2}} \right]}
\\
\\
\\
\\
\frac{dy}{dx} =& \frac{\displaystyle \frac{2a (b^2)}{b^2 + a^2}}{\displaystyle b - \frac{2b(a^2)}{a^2 + b^2}}
\\
\\
\\
\\
\frac{dy}{dx} =& \frac{\displaystyle \frac{2ab^2}{\cancel{b^2 + a^2}}}{\displaystyle \frac{a^2 b + b^3 - 2ba^2}{\cancel{a^2 + b^2}}}
\\
\\
\frac{dy}{dx} =& \frac{2ab^2}{a^2 b + b^3 - 2a^2 b} = \frac{b(2ab)}{b(a^2 + b^2 - 2a^2)}
\\
\\
\frac{dy}{dx} =& \frac{2ab}{a^2 + b^2 - 2a^2} = \frac{2ab}{-a^2 + b^2}

\end{aligned}
\end{equation}
$



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..

$\displaystyle \left( \frac{a^2 - b^2}{2ab} \right) \left( \frac{2ab}{-a^2 + b^2} \right) \longrightarrow \left( \frac{a^2 - b^2}{2ab} \right) \left( \frac{-2ab}{(a^2 - b^2)} \right) = -1$

Therefore, the curve must be orthagonal.