Prove that 1/x is a hyperbola 
Prove that 1/x is a hyperbola with foci given by $(\sqrt2,\sqrt2)$
  and $(-\sqrt2,-\sqrt2)$.

The idea is to use ONLY the definition of a hyperbola, that is:

Let $F_1$ and $F_2$ be two given points.  A hyperbola $H$ is the set of points in the plane such that $$|dist(P,F_1)-dist(P,F_2)| =|dist(Q,F_1)-dist(Q,F_2)|,$$ for all $P,Q$ in $H$.

To be more precise, I don't want to use any notion of matrices since this is having no background on linear algebra.
What I did was to pick a generic point of the form $(x,1/x)$ and work with $$|dist(F_1,P)-dist(F_2,P)|,$$ where $F_1=(\sqrt2,\sqrt2), F_2=(-\sqrt2,-\sqrt2)$ and $P=(x,1/x)$.   The problem was that I cannot prove that the result is independent of $x$.
Since that didn't work and knowing that $(1,1)$ belongs to the hyperbola, I calculate the difference of the distances between that point and the foci and the result is
$$dist(F_2,(1,1))-dist(F_1,(1,1))=\sqrt2(\sqrt2+1)- \sqrt2(\sqrt2-1)=2\sqrt2.$$
Therefore, the constant value that I am expecting to get is $2\sqrt2$, which I can't get in the general case!
After that, I tried with $(2,1/2)$, the result was 
$$dist(F_2,(2,1/2))-dist(F_1,(2,1/2))=\frac{\sqrt{33+20\sqrt2}}{2}-\frac{\sqrt{33-20\sqrt2}}{2} $$
The key is to notice that  $$\frac{\sqrt{33\pm 20\sqrt2}}{2}= \frac{5}{2}\pm \sqrt2.$$  (that is according to Wolfram).  However, I don't know how to make an algebraic arrangement in order to get that.  With that in mind, I am hoping to get for the general case something like $$f(x)+\sqrt2-(f(x)-\sqrt2),$$ where of course $f(x)$ should be some number depending on $x$.  However, I am not being able to do it.
Any help is appreciated.
Thanks.
 A: Note that since $F_i$ are points, 
$$ dist(F_i,P) = |F_i - P|.$$
Setting $F = F_1=(\sqrt2,\sqrt2)$, we have $-F=F_2$. Then we are asked to check that
$$ \Big||P-F| - |P+F|\Big| \text{ is constant in $P$,}$$
where $P=(x,1/x)$ is an arbitrary point in the graph of $1/x$. We compute directly,
$$ |P\pm F|^2 = (x\pm \sqrt 2)^2 + (1/x \pm \sqrt2)^2 = x^2 + x^{-2} +4 \pm 2\sqrt{2}(x+x^{-1})  $$
The following is perhaps a trick, 
$$ x^2 + x^{-2} + 2 = (x+x^{-1})^2$$
Recognizing the importance of $x+x^{-1}$, let $t = x+x^{-1}$. Note that for $x\in\mathbb R\setminus \{0\}$, $|t|\ge \color{orange}{2}$. Then
$$ |P\pm F|^2 = t^2 \pm 2\sqrt 2 t + (\sqrt{2})^2 = (t \pm \sqrt 2)^2$$
Therefore $$ |P+ F| - |P- F| = |t + \sqrt 2| - |t - \sqrt 2| =
 \begin{cases}
2\sqrt{2}&t>\color{blue}{\sqrt{2}},\\[0.6em]
-2\sqrt{2} & t<-\color{blue}{\sqrt{2}},\\
\text{who cares} & \text{otherwise.}
\end{cases}$$
Since $\color{blue}{\sqrt{2}}<\color{orange}2$,  $\Big||P+ F| - |P- F|\Big| = 2\sqrt{2}$ for all $P$ of the form $(x,1/x)$, where $x\in \mathbb R\setminus \{0\} $.
