# Thread: Orthogonal Trajectories

1. ## Orthogonal Trajectories

"Find the orthogonal trajectories of the family of curves $x^2+y^2=2cx$.

I know the steps: 1) solve for c...okay... $c=(x^2+y^2)/2x$.
2)Find y', and plug the c back into it.
3) set dy/dx equal to the negative reciprocal of y' from pt. 2, and solve for y. This is the family of orth. traj.

My problem is finding y' for part 2. Well, I can find y', but getting it into a form so that I can solve for y in part 3.

$y=±(2cx-x^2)^(1/2)$

y'= $1/(2sqrt((x^2+y^2)/x-x^2)*((x^2+y^2)/x-2x)$

Where should I go from here? Thanks!

2. Originally Posted by lakesfan210
"Find the orthogonal trajectories of the family of curves $x^2+y^2=2cx$.

I know the steps: 1) solve for c...okay... $c=(x^2+y^2)/2x$.
2)Find y', and plug the c back into it.
3) set dy/dx equal to the negative reciprocal of y' from pt. 2, and solve for y. This is the family of orth. traj.

My problem is finding y' for part 2. Well, I can find y', but getting it into a form so that I can solve for y in part 3.

$y=±(2cx-x^2)^(1/2)$

y'= $1/(2sqrt((x^2+y^2)/x-x^2)*((x^2+y^2)/x-2x)$

Where should I go from here? Thanks!
Have you been taught implicit differentiation:

$2x + 2y \frac{dy}{dx} = 2c \Rightarrow \frac{dy}{dx} = \frac{c - x}{y}$.