Orthogonal Trajectories of Cassinian Curves Preamble
The Cassinian curves are the pre-images of concentric circles (centered at $1+0\,i$) under the map $z\mapsto z^2$. Using this fact and the fact that complex polynomials are conformal we can deduce that the orthogonal trajectories to the Cassinian curves map to straight lines passing through the point $1+0\,i$. These lines can be writen as
$$w(\lambda) = 1 + \lambda \,e^{i\theta}$$
Where $\theta$ is the angle the line makes with the real axis. Applying the function $z\mapsto \sqrt{z}$ to these lines then gives us back the othogonal trajectories to the Cassinian curves.
Question
The book I am working through asks  the reader to use this to show that the orthogonal trajectories are hyperbolae but I can't seem to make much headway with it. How does one show this?
Attempt
I tried expanding the real and imaginary components of the line ($w=u+i\,v$) and its image ($z=x+i\,y$), equating $z=\sqrt{w}$ and squaring both sides, which gives
$$x^2-y^2 + 2 i\,xy = \underbrace{1+\lambda\,\cos(\theta)}_u + i\,\underbrace{\lambda\sin(\theta)}_v$$
but apart from the fact we have $x^2-y^2$ in there this doesn't seem so useful to me. It might also be useful to note that the orthogonal trajectories must pass through $z=\pm 1$.
 A: The preimage of the real line under $f(z) = z^2$ is not a hyperbola, it's the limiting case, the two coordinate axes.
For all other straight lines passing through $1$, let us describe them by an equation: $L_c = \{ u+i v : u-1 = cv\}$, and let $H_c = f^{-1}(L_c)$. For $H_c$, we then obtain the describing equation
$$x^2-y^2-1 = c(2xy)\qquad \text{resp.} \qquad x^2 - 2cxy - y^2 = 1.\tag{1}$$
That is the equation of a hyperbola, but not yet in standard form (except for $c = 0$). We obtain an equation in standard form by rotating the coordinate system.
Let $\alpha = \frac{1}{2} \arctan c$. Then we can write $(1)$ as
$$\begin{align}
x^2 - 2xy\tan 2\alpha - y^2 &= 1\\
x^2\cos 2\alpha - 2xy\sin 2\alpha - y^2\cos 2\alpha &= \cos 2\alpha\\
(x\cos\alpha - y\sin\alpha)^2 - (x\sin\alpha + y\cos\alpha)^2 &= \cos 2\alpha
\end{align}$$
and in the rotated coordinates
$$\begin{pmatrix}\xi\\ \eta\end{pmatrix} = \begin{pmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha &\cos\alpha \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix}$$
we have the standard form
$$\xi^2 - \eta^2 = \cos 2\alpha.\tag{2}$$
Thus we obtain
$$H_{c} = e^{\large -\frac{i}{2}\arctan c}\cdot \left\{\xi + i\eta : \xi^2-\eta^2 = \frac{1}{\sqrt{1+c^2}}\right\}.$$
A: Too late but I thought may be this naive approach is worth writing here.
Preimage of the concentric circles under the mapping $z \mapsto z^2$ form the Cassinian curves with focii $\pm 1 + 0i$ (preimages of the centre of the circles). The equation of the Cassinian curves for some positive constant $k$ will then be,
$$|z-1||z+1| = k \qquad \equiv \qquad |z-1|^2|z+1|^2 = k^2$$
Representing in Cartesian coordinates,
$$\begin{align*}
((x-1)^2+y^2)((x+1)^2+y^2) &= k^2\\
(x^2-1)^2 + y^4 + 2y^2(x^2+1) &= k^2
\end{align*}$$
Implicit differentiation,
$$x(x^2-1)dx + y^3dy +  xy^2dx + x^2ydy + ydy = 0$$
To obtain differential equation corresponding to the orthogonal trajectory, replace $dy/dx$ by $-dy/dx$ to get,
$$x(x^2-1)dy -y^3dx+xy^2dy -x^2ydx -ydx = 0$$
Rearraging terms,
$$\begin{align*}
y^2(xdy-ydx) - x^2(ydx-xdy) &= xdy + ydx\\
x^2y^2 d(y/x) - x^2y^2 d(x/y) &= d(yx)\\
d(y/x)-d(x/y) &= (d(yx))/((xy)^2)
\end{align*}$$
Indefinite integration to get,
$$\begin{align*}
y/x - x/y &= -1/(xy) + c\\
x^2-y^2 +cxy &= 1
\end{align*}$$
