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

Question

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.

Answer 1

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 $PQ,$

$$2u=2t+h\iff h=?,2(u+1)=2/t+k\iff k=?$$

$$2u-2t+2(u+1)-2/t=2t+2/t\iff2u=2t+2/t-1$$

$$\implies h=2u-2t=2/t-1\text{ and } k=2u+2-2/t=1+2t$$

$$\implies(h+1)(k-1)=2/t\cdot2t=4$$

Answer 2

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 $(x+1)(y-1)=4$.

Answer 3

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, $$ xy=x-y+5 $$