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 dxdy=f(x,y).
The orthogonal trajectories then are the solutions of dxdy=−f(x,y)1.
For many families of curves, one cannot explicitly solve for dxdy and obtain a differential equation of the form dxdy=f(x,y).
Example 1. Find the orthogonal trajectories of the family of curves x2+y2=c2.
Rewrite as x2+y2−c2=0, or F(x,y,c)=x2+y2−c2=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+2ydxdy=0, or dxdy=−yx.
Here, f(x,y)=−yx; so, the orthogonal trajectories satisfy the equation dxdy=xy. This equation is separable: ydy=xdx, or ln(y)=ln(x)+a, which can be rewritten as y=kx, where k=ea.
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 y2=4cx.
Rewrite as y2−4cx=0, or F(x,y,c)=y2−4cx=0. Implicitly differentiating the given equation with respect to x, we obtain that 2ydxdy−4c=0, or dxdy=y2c.
Note that, from the initial equation, c=4xy2; so, dxdy=y2(4xy2), or dxdy=2xy.
Here, f(x,y)=2xy; so, the orthogonal trajectories satisfy the equation dxdy=−2yx. This equation is separable: it can be rewritten as ydy=−2xdx, or 2y2=−x2+k.
So, the orthogonal trajectories are x2+2y2=k (k>0).