Prove that the graphic of $\frac 1 x$ is a branch of hyperbola

Question

Let $f:\Bbb R_+\to\Bbb R_+,x \mapsto y=f(x)=1/x$ Prove that the graphic of $f$ is a branch of hyperbola.
Rotating by $\frac \pi 4$ with the matrix $M_-= \pmatrix {cos\alpha &&sin\alpha \\sin\alpha&&-cos\alpha}$. ie $$\pmatrix{x\\y}=\pmatrix{\frac {\sqrt2} 2&&\frac {\sqrt2} 2\\\frac {\sqrt2} 2&&-\frac {\sqrt2} 2}\pmatrix{x'\\y'}$$ we get $$x=\frac {\sqrt2} 2(x'+y'),y=\frac {\sqrt2} 2(x'-y'),$$ $$x\ge0\iff x'+y'\ge0, y>0 \iff x'-y'\ge0$$ then $f$ can be written as $$\frac {x'^2} 2 - \frac {y'^2} 2 = 1, x'\ge |y'| $$
how do I proceed?

Answer 1

Using parametrization of one branch:

\begin{align} (x',y') &= \sqrt{2}(\cosh t,\sinh t) \\ (x,y) &= \frac{(x'+y',x'-y')}{\sqrt{2}} \\ &= (\cosh t+\sinh t,\cosh t-\sinh t) \\ &= (e^{t},e^{-t}) \end{align}