Given $f:\left(x,y\right) \mapsto \left(u,v\right)$, to find regions in $xy$-plane that map onto a given region in $uv$-plane

Question

Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be given by $$\left(u,v\right)=f\left(x,y\right)=\left(x-y,xy\right)$$ Problem : Which regions in $xy$-plane map onto the rectangle $\left[0,1\right] \times \left[1,4\right]$ in the $uv$-plane.

My progress : $u=x-y,\,v=xy \Rightarrow y=\frac{-u\pm\sqrt{u^2+4v}}{2},\,x=\frac{2v}{-u\pm\sqrt{u^2+4v}}$. Thus, it is clear that there will be two regions in the $xy$-plane (one corresponding to $+\sqrt{u^2+4v}$ and the other corresponding to $-\sqrt{u^2+4v}$) that map to $\left[0,1\right] \times \left[1,4\right]$ in the $uv$-plane. The regions will be given by \begin{align*} &R_1=\left\{\left(\frac{2v}{-u+\sqrt{u^2+4v}},\frac{-u+\sqrt{u^2+4v}}{2}\right):\left(u,v\right) \in \left[0,1\right] \times \left[1,4\right]\right\}\\ &R_2=\left\{\left(\frac{2v}{-u-\sqrt{u^2+4v}},\frac{-u-\sqrt{u^2+4v}}{2}\right):\left(u,v\right) \in \left[0,1\right] \times \left[1,4\right]\right\} \end{align*}

Where I am stuck : I cannot find $R_1$ and $R_2$ explicitly. Any help regarding this would be much appreciated.

Answer 1

Hint: Here is a plot:

\begin{align}\color{blue}{0\leq x-y\leq1}\\ \color{green}{1\leq xy\leq4}\\ \end{align}

A solution could thus be: $$(x,y)\in[1,2]\times \{1\}\cup\{2\}\times [1,2]$$ In general, any region that contains a curve which goes from one boundary of the blue region to the other, and the same for the green curve will do.

Answer 2

HINT

The segment from $(0,1)$ to $(1,1)$ that is $(x,y)=(t,1)$ with $t\in[0,1]$ is mapped to

$$(u,v)=(x-y,xy)=(t-1,t) \quad v=u+1$$

from $(-1,0)$ to $(0,1)$ and so on.