Question

the orthogonal trajectories of x^2+4y^2=a^2​

Answer 1

Let

φλ(x,y)=x2a2+λ+y2b2+λ

and (u,v) be a point lying on the intersection of the curves φλ1(x,y)=1 and φλ2(x,y)=1.

The normal for the ith curve at (u,v) is in the direction

∇φλi(x,y)|(u,v)=(2ua2+λi,2vb2+λi)

The dot product of these two normal vectors is proportional to

[∇φλ1(x,y)⋅∇φλ2(x,y)](u,v)=4u2(a2+λ1)(a2+λ2)+4v2(b2+λ1)(b2+λ2)=4λ2−λ1[(u2a2+λ1+v2b2+λ1)−(u2a2+λ2+v2b2+λ2)]=4λ2−λ1(1−1)=0