Answer to Question #93548 in Differential Equations for Mahesh patro

Question

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

Answer 1

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

$y=\sqrt{\frac xc} \quad \perp \quad Y=Y(x), \quad x>0, \quad y>0.$

Then

$Y'=-\frac{1}{y'}=-2\sqrt{cx}.$

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

$Y'=-\frac{2x}{Y} \quad \text{or} \quad Y^2+2x^2=C.$

We set the family of ellipses

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