Is there a way to find out the fixed curve to which variable lines are normal to? Recently I came across a question, which is as follows: 
For all real values of $k$, what is the fixed curve the line 
$$
y = (x - 11) \cos(k) - \cos (3k) 
$$
is normal to.
This could be done easily by rewriting the equation is terms of cos(k)
which yielded the form: 
$$y= mx - 2am - am^3,$$ 
which is the equation of normal to the standard parabola.
But on a more general note, there must be way to find the curve to which the variable line is normal to. The tricky part for me is, the constant term also is variable in this case.
For a very rudimentary case, suppose 
$$y= mx$$
I did the following:
\begin{align}
    y=  mx\\
    m=  y/x\\
    0= (y'x - y)/x^2 \\
    y'x= y\\
    y dy= - xdx\\
    x^2 + y^2= 2c
\end{align}
Which we all know is the equation of a standard circle .  
But for a general problem as such, I tried multiple approaches but invariably got stuck at some point or the other.
 A: Your question is about orthogonal trajectories of a curve that must be set within a single unifying DE. You need to take negative reciprocal of derivative as per standard procedure and re-integrate. 
The variable curves are right opening con-focal parabolas of focal length $f=1$ having equation
$$ y^2= 4 f(x+f)$$
whose differential equation is obtained by once differentiating and eliminating $f$ as
$$ y^2=2 x y y'+ y^2 y^{'2}$$
$$ y'\rightarrow \frac {-1}{y^{'}} $$
By substitution the new DE of orthogonal trajectories is
$$ y^2 =2 x y y'+ y^2 y^{'2} $$
which is unaltered in this particular case. (These arise btw as real/imaginary parts of complex function $ w= z^2$).

By integration and choosing a point on it as initial condition a curve marked as red is selected as your fixed curve orthogonal to all given right opening parabolas ignoring the rest.You can look at any one of them associating a corresponding arbitrary constant and ignore others that you do not want to see.
$$ y^2= 4 f(-x+f)$$
