I want to find a parametric curve that would be perpendicular to all curves $y=c \sin x$
I can see that these curves will be straight lines when $x=\frac{2n+1}2\pi$ and they should become tiny circles as $x\rightarrow n\pi$, but I do not see how I would do this mathematically or what the answer would look like.
The only thing that came to my mind so far is to think of a function $F$ and equate its derivative to be $-\frac{1}{c \cos x}$
$$-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}=-\frac{1}{c \cos x}$$
$$c\cos x\frac{\partial F}{\partial x}=\frac{\partial F}{\partial y}$$
A separable solution to this would be
$$\exp \bigg(\frac{2 k \tanh ^{-1}\left(\tan \left(\frac{x}{2}\right)\right)}{c}+ky\bigg)$$
I want the parametric curves to be perpendicular for all $c$. But I do not know if this makes sense and how I would continue.
Intuitively, I am expecting to see concentric ellipses centered at $x=n\pi$
The original family is: $$ g(x, y) = \frac{y}{\sin(x)} = c $$ Implicit differentiation regarding $x$ gives: $$ \frac{y'\, \sin(x) - y \, \cos(x)}{\sin(x)^2} = 0 \Rightarrow \\ y' = \cot(x) \, y $$ An orthogonal solution must satisfy the ODE: $$ y' = - \tan(x) \, \frac{1}{y} $$ Separation of variables leads to $$ \int\! y\, dy = -\! \int\! \tan(x)\,dx \Rightarrow \\ \frac{1}{2} y^2 = \ln(\cos(x)) + C \Rightarrow \\ y = \pm \sqrt{2 \ln(\cos(x)) + d}) $$ for some integration constant $d = 2C$.
As Rahul noted, to get all solutions one must use: $$ y = \pm \sqrt{2 \ln(\lvert\cos(x)\rvert) + d}) $$ Looks like this issue.
Here is a first image:
Here is another one: