prove $x^2 = 4 c(y+c)$ is self orthogonal trajectory Can someone come up with a proof with this $$x^2 = 4 c(y+c)$$ to be self orthogonal?
I know in have to put $-dx/dy$ instead of $dy/dx$ but I cant solve it
 A: Suppose there are two curves with parameters $c_1$ and $c_2$. Then at their points $(x,y)$ of intersection, $x^2 = 4c_1(y+c_1)$ and $x^2 = 4c_2 (y+c_2)$. These can be easily solved, giving $y = -c_1 - c_2$ and $x^2 = -4c_1c_2$.
For the curves to be orthogonal at a point, they must satisfy
$$
\left(\frac{dy}{dx}\right)_1\left(\frac{dy}{dx}\right)_2 = -1.
$$
Differentiating the defining equation with respect to $x$ gives $dy/dx = x/(2c)$, so we have
$$
\left(\frac{dy}{dx}\right)_1\left(\frac{dy}{dx}\right)_2 =\left(\frac{x}{2c_1}\right)\left(\frac{x}{2c_2}\right) = \frac{x^2}{4c_1c_2} =  -1,
$$
as desired.
A: Let us split the unique family of curves (parabolas with $oy$ as symmetry axes) into two families. 
See figure below, borrowed to (http://xahlee.info/SpecialPlaneCurves_dir/ConicSections_dir/conicSections.html):
$$\tag{1}\begin{cases}F_c(x,y)&=&x^2-4c(y+c)&=&0 \ \text{with} \ c>0 \ \text{(green parabolas)}\\G_{c'}(x,y)&=&x^2+4c'(y-c')&=&0 \ \text{with} \ c'>0  \ \text{(purple parabolas)}\end{cases}$$
Solving system (1), it is easy to obtain the coordinates of generic intersection point(s) $I$ which are:
$$\tag{2}x=\pm \sqrt{4cc'}, \ \ y=c'-c.$$
It suffices to prove that the normal vectors at point(s) $I$ to the $F_c$ curve and to the  $G_{c'}$ curve (featured in black) are orthogonal.
It is easy because the normal vectors are given by the gradient operation :
$$\binom{\tfrac{\partial F_c}{\partial x}}{\tfrac{\partial F_c}{\partial y}}=\binom{\ \ 2x}{-4c} \ \ \text{and} \ \ \binom{\tfrac{\partial G_{c'}}{\partial x}}{\tfrac{\partial G_{c'}}{\partial y}}=\binom{2x}{4c'}$$
The dot product of these normal vectors,
$$4x^2-16cc',$$
is clearly $0$ when  $x=\pm \sqrt{2cc'}$ (abscissas of points $I$ (see (2))).
Remarks: 
1) It can be shown that all these parabolas have a common focus: the origin.
2) These curves are connected to complex function $f(z)=\sqrt{z}$ : see for example homofocal parabolas here

A: The ODE for the curves of the family $$ x^2= 4c(y+c)~~~(1) $$ Differentiating w.r.t. $x$, we get $2x=4cy'$ putting $c=\frac{x}{2y'}$ in (1) we get the ODE for (1)
$$x^2=\frac{2xy}{y'}+\frac{x^2}{y'^2} \implies x^2y'^2=2xyy'+x^2~~~(2)$$
Now if we change $y'$ to $-\frac{1}{y'}$ in (2), we get the ODE for the orthogonal trajectories to (1) as
$$\frac{x^2}{y'^2}=-\frac{2xy}{y'}+x^2  \implies x^2y'^2=2xyy'+x^2~~~(3)$$
Eq. (2) and (3) being identical, the self orthogonality of (1) is proved.
