[Math] Orthogonal Trajectories and Parameters

ordinary differential equations

What's the difference between constants and parameters in terms of finding orthogonal trajectories to a given family of curves?

I understand that for a curve like
$x^2 + y^2 + 2gx + c = 0$,
the process of finding an orthogonal trajectory involves finding the gradient $dy/dx$ in terms of y and x only. We can then use this expression to calculate the orthogonal trajectory's gradient (the negative reciprocal of $dy/dx$) and solve it to get a function of y in terms of x.

My textbook says g is a parameter but says nothing about c. Does this mean g will appear in the expression for the orthogonal trajectory? If so, is this because a parameter in the equation of a family of curves affects the shape of the orthogonal trajectory and hence must appear in the expression for the latter?

I've worked out the orthogonal trajectory to be
$y = k(g + x)$ using this method. Please let me know if and where I'm making a mistake.

Best Answer

You are doing it right.

$$x^2+y^2+2gx+c=0$$

$$x^2+2gx+(g)^2-g^2+y^2+c=0$$

$$(x+g)^2+y^2=g^2-c$$

There you go the equation of a typical circle. centered at $(-g,0)$

The radius of the circle is then $\sqrt{g^2+c}$

Finding orthogonal trajectory, we have three steps.

Step 1: Differentiating

$$2(x+g)+2y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x+g}{y}$$

Step 2: Taking negative reciprocal of the differential equation

$$\frac{dy}{dx}=\frac{y}{x+g}$$

Step 3: Solving the differential equation

$$\frac{dy}{dx}=\frac{y}{x+g}$$

$$\frac{dy}{y}=\frac{dx}{x+g}$$

$$\int \frac{dy}{y}=\int \frac{dx}{x+g} $$

$$\ln|y|=\ln|x+g|+k$$

$$y=k(x+g)$$

This is an equation of a straight line.

It will look something like this,

enter image description here

Notice that the linear equations always has a root at $x=-g$ which is also the center of the circle equation. You are free to choose which center $g$.