I stumbled with this problem in a notebook that has been bothering for the whole day(actually 3)...The answer is written but there's no explanation nor a steb-by-step procedure or anything. If you know how to analyse the problem and create an equation from this, I will be very grateful!
Find the ortogonal trajectories at the cissoid :
$(2a-x) y^2 = x^3$
Answer: $(x^2 + y^2) = C(2x^2 + 2y^2)$
Thank you for your help!
if you difference $(2a - x)y^2 = x^3$ you get the tangent to the cissoid satisfies $$(2a-x)2y\, dy -(y^2 + 3x^2) \, dx = 0 .$$ therefore, the normal to the cissoid satisfies the differential equation $$(2a-x)2y\, dy + (y^2 + 3x^2) \, dx = 0 .$$
we can write it as a system two differential equations $$\frac{dx}{dt} = x - 2a, \frac{d(y^2)}{dt} = y^2 + 3x^2.$$ we can solve these as $$x = 2a+ Ce^t, y^2 = De^t + 3\left(4a^2 + 2aC\, te^t + C^2 e^{2t}\right)$$
we can eliminate the $t$ between the $x$ and $y$ to get the family of normal curves to the cissoid
$$ y^2 = \frac{D}{C}(x-2a) + 3\left(4a^2 + 2a(x-2a)\ln(\frac{x-2a}{C}) + (x-2a)^2\right) $$
this is much more more complicated than the one you have. i hope i have not made any errors.