# Orthogonal Trajectories

Consider a one-parameter family of curves in the xy-plane defined by F(x,y,c)=0 where c denotes the parameter. The problem is to find another one-parameter family of curves, called the orthogonal trajectories of the family defined above and given analytically by G(x,y,k)=0 such that every curve in this new family intersects at right angles every curve in the original family.

We implicitly differentiate first equation with respect to x, then eliminate c between this derived equation and equation F(x,y,c)=0. This gives an equation connecting x, y, and y', which we solve for y' to obtain a differential equation of the form (dy)/(dx)=f(x,y) .

The orthogonal trajectories then are the solutions of (dy)/(dx)=-1/(f(x,y)) .

For many families of curves, one cannot explicitly solve for (dy)/(dx) and obtain a differential equation of the form (dy)/(dx)=f(x,y) .

Example 1. Find the orthogonal trajectories of the family of curves x^2+y^2=c^2 .

Rewrite as x^2+y^2-c^2= or F(x,y,c)=x^2+y^2-c^2=0 . This family of curves is consists of circles with centers at the origin and radii c. Implicitly differentiating the given equation with respect to x, we obtain that 2x+2y(dy)/(dx)=0 or (dy)/(dx)=-x/y.

Here f(x,y)=-x/y so orthogonal trajectories satisfy equation (dy)/(dx)=y/x . This equation is separable: (dy)/y=(dx)/x or ln(y)=ln(x)+a which can be rewritten as y=kx where k=e^a .

Solution y=kx represents orthogonal trajectories (straight lines that pass through the origin).

Example 2. Find the orthogonal trajectories of the family of curves y^2=4cx .

Rewrite as y^2-4cx=0 or F(x,y,c)=y^2-4cx=0 . Implicitly differentiating the given equation with respect to x, we obtain that 2y(dy)/(dx)-4c=0 or (dy)/(dx)=(2c)/y .

Note that from initial equation c=(y^2)/(4x) , so (dy)/(dx)=(2((y^2)/(4x)))/y or (dy)/(dx)=y/(2x) .

Here f(x,y)=y/(2x) so orthogonal trajectories satisfy equation (dy)/(dx)=-2x/y . This equation is separable: it can be rewritten as ydy=-2xdx or (y^2)/2=-x^2+k .

So, orthogonal trajectories are x^2+(y^2)/2=k (k>0).