# 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).