Question

Find the orthogonal trajectories of the family of parabolas x = cy²

Answer 1

Let $(x,y=Y)$ be the intersection point of the curves

$y=\sqrt{x/c} \quad \bot \quad Y=Y(x), \quad x>0, \quad y>0.$

Then

$Y'=-1/y' \quad \lrArr \quad Y'=-2\sqrt {cx}.$

But $c=\frac{x}{y^2}.$ Therefore

$Y'=-\frac{2x}{Y} \quad \text{or} \quad YdY=-2xdx \quad \text{or} \quad \int YdY=-\int 2xdx.$

Then

$Y^2+2x^2=C$

We get the family of ellipses finally:

$\frac{x^2}{a^2} + \frac{y^2}{2a^2}=1$.