Orthogonal trajectories - why is it necessary to isolate the parameter For orthogonal trajectory, I realized that I need to express the parameter of the given family of curves in terms of x and y, in order to get the right answer. 
e.g. in $y = kx$, $k$ is the parameter I was talking about in the preceding sentence above.
1) If I solve the orthogonal trajectory problem treating the parameter as if it is a constant. What will that constant mean in my final answer?
2) If I solve the problem properly (i.e. express $k$ in terms of $x$ and $y$), does $k$ changes as I move along any one of the orthogonal trajectories? Is this why I need to isolate for $k$ in the first place?
Thanks for the help
 A: You are confusing your particular problem ($y=kx$) for the general orthogonal trajectory problem.
The curves you are searching for must intersect each member of the given family of curves at a right angle. That means the key point is the slope of the tangent line at each point on each curve in your family. Therefore, for each $k$ you are interested in $\frac{dy}{dx}$, and you want to put it in terms of $x$ and $y$ with no $k$ in it. In some problems you may solve for $k$ to more easily eliminate it, but that will not always be the case.
In your particular problem, $\frac{dy}{dx}$ equals $k$. So you are interested in the constant in this particular problem, but only because it happens to equal the curve's slope. You solve for $k$ so you can find the slope in terms of $x$ and $y$ without $k$ (in this case, $\frac{dy}{dx}=\frac yx$).
So the answers to your questions are:


*

*You treat the parameter as a constant because for a given curve in the family of curves it is a constant, and only $x$ and $y$ change. You use these changes to find $\frac{dy}{dx}$ in any form then to get it as a function of $x$ and $y$ without the parameter. The constant itself is meaningless in your answer at this point. You have eliminated it, after all. The meaning of the constant overall depends on the particular problem, and is just the way to single out one of the curves from the rest in the family of curves. In your particular case it means the slope of the curve, which is constant in your particular case. In other problems it will have other meanings.

*As "move along any one of the orthogonal trajectories" the parameter for the curve in the original family of curves indeed changes. Each point on your orthogonal trajectory will have a different parameter value in the original family of curves. You can see this must be the case: your new curve is perpendicular to the original curve, so as soon as the point moves on the orthogonal curve away from its initial point it moves perpendicular to the original curve and thus moves off that curve. It now moves onto other curves in the family, each with its own parameter value. However, this changing $k$ is not the reason your solved your original equation for $k$, as I have explained.


In your particular problem, the original family has the slopes
$$\frac{dy}{dx}=\frac yx$$
so the orthogonal family has the negative-reciprocal slopes
$$\frac{dy}{dx}=-\frac xy$$
and solving that (using separation of variables) gives
$$x^2+y^2=C$$
where $C$ is the parameter for your orthogonal family of curves. In this problem the new parameter has the meaning "square of the radius of this circle" since each orthogonal curve is a circle. But a different problem will have a different meaning for $k$ and for $C$.
