If the reflection of the hyperbola $xy = 4$ in the line $x - y + 1 = 0$ is $xy = mx + ny + l$ find $m + n + l$ 
If the reflection of the hyperbola $xy = 4$ in the line $x - y + 1 = 0$ is $xy = mx + ny + l$ find $m + n + l$.

I already solved it by taking a general point (more than one way possible for it) and then finding the reflection and then substituting it in the equation but I found that very lengthy. Besides there are more than one way to find the reflection of a point on a line. 
Please give the simplest solution possible for this problem even if it is the same as my method.
 A: Any point on $xy=4$ will be $P(2t,2/t)$
Now any point on $L:x-y+1=0$ can be written as $Q(u,u+1)$
If $R(h,k)$ is any point on $$xy=mx+ny+l,$$  
$Q$ will be the perpendicular bisector of $PR$
The equation of $PQ:$ $$x+y=2t+2/t\implies h+k=2t+2/t$$
Also $Q$ being the  midpoint of $PQ,$
$$2u=2t+h\iff h=?,2(u+1)=2/t+k\iff k=?$$
$$2u-2t+2(u+1)-2/t=2t+2/t\iff2u=2t+2/t-1$$
$$\implies h=2u-2t=2/t-1\text{ and } k=2u+2-2/t=1+2t$$
$$\implies(h+1)(k-1)=2/t\cdot2t=4$$
A: Because of the symmetry of the hyperbola, reflecting it over a diagonal line (by which I mean a line of the form $y=\pm x +a$) is the same as translating it. The centre of symmetry of the original hyperbola (the origin) is moved to $(-1,1)$, which means that the new hyperbola is given by the equation $(x+1)(y-1)=4$.
A: Change coordinates, setting $X=x+1$ and $Y=y$. Then the line of reflection is $X-Y=0$, so the reflection is just exchanging coordinates.
The hyperbola $xy=4$ becomes $(X-1)Y=0$, so its transformed is $(Y-1)X=4$. Returning to the original coordinate system you get
$$
(y-1)(x+1)=4
$$
that is,
$$
xy=x-y+5
$$
