Closed form of the orthogonal trajectories to the lines: $y=mx+ \sqrt{1+m^2}$ Normally, the discussions/ exercises of orthogonal trajectories (OTs) to a family of lines are  avoided in textbooks, with $y=mx$ or  $mx+m$ being the exceptions where one gets a family of circles. For other family of lines which involve a parameter say $m$ in non-linear ways, finding OTs could be an involved calculation wherein first order non-linear ODEs are encountered whose solvability cannot be guaranteed. 
Recently, we required a closed or parametric form of the OTs for the family of lines $$y=mx+\sqrt{1+m^2}~~~~(1)$$ We eliminate the parametre $m$ by differentiating w.r.t. $x$ to get a first-order ODE. Next, we change $y'$ to $-1/y'$, then by setting $y'=p$ we solve the  interesting ensuing ODE. We find a parametric form for the required OTs in terms of $x(p)$ and $y(p)$  a free constant say $C$.
Interestingly, our expressions for the required OTs succeed but not without displaying an unwarranted feature!
We therefore need your results of the  OTs for (1), please help. We may post our results later  
 A: $$y=mx+\sqrt{1+m^2}~~~~~(1)$$
Differentiating w.r.t. $x$ to get $m=y'$, by changing $y'$ to $-1/y'$. denoting $y'$ by $p$ we get the ist order 
non-linear ODE for the orthogonal family of curves (OFC) to (1) as
$$yp=-x+\sqrt{1+p^2},~~ p=\frac{dy}{dx}~~~~(2)$$
Differentiating (2) w.r.t $y$ we get
$$\left( py-\frac{p^2}{\sqrt{1+p^2}} \right) dp + (1+p^2) dy=0 ~~~~(3)$$
This is of the type $M(p,y)dp+N(p,y)dy=0$ which is not-exact as $\frac{\partial M}{\partial y}=p$ and $\frac{\partial N}{\partial p}=2p$ are unequal. For its integrating factor $I$
$$\mu(p,y)=\frac{1}{N} \left(\frac{\partial M}{\partial y}- \frac{\partial N}{\partial p} \right)=\frac{-p}{\sqrt{1+p^2}} \implies I=e^{\mu}=\frac{1}{\sqrt{1+p^2}} ~~~~(4)$$
Multiplying by $I$, (3) becomes 
$$\left( \frac{py}{\sqrt{1+p^2}}-\frac{p^2}{1+p^2} \right) dp + \sqrt{1+p^2} dy =0 ~~~(5)$$
exact whose solution is given by
$$\int \left( \frac{py}{\sqrt{1+p^2}}-\frac{p^2}{1+p^2} \right) dp ~[\mbox {treat $y$ as constant}] =C ~~~~(6)$$ We get
$$y(p)=\frac{C+ p-\tan^{-1} p}{\sqrt{1+p^2}},~~~\mbox{by Eq. (2)}~~ x(p)= \sqrt{1+p^2}-p y(p) ~~~~~~~(7)$$
where $p$ is treated as a real parameter giving us the parametric form of OFC for (1).
In the following, we present  the plots of OFCs  (7) and lines (1) together. 
We take $C=\pm 10, ~\pm 12, ~\pm 13,~ \pm 14$. The lower parts of the OFCs are for $C<0$. For lines we choose m$m=\pm 1, \pm 2$.
In  Fig. (a) all the lines are cutting the OFCs orthogonally, the lines of the types $y=m+\sqrt{1+m^2}, m<0$ or $y=mx -\sqrt{1+m^2}, m>0$.
But in Fig.(b) the same OFCs but the lines are cutting them non-orthogonally.
The lines are of the types: $y=mx+\sqrt{1+n^2}, m>0$, $y=mx-\sqrt{1+m^2}, m<0$. Also theis (b) category of lines may not cut the OFCs otained in Eq. (7).


