# 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 the first equation with respect to x, then eliminate c between this derived equation and the 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=0, or F(x,y,c)=x^2+y^2-c^2=0. This family of curves 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, the orthogonal trajectories satisfy the 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.

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

Let's solve another example.

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

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