Change of variables problem. 
We're given a normal xy - plane , with the help of the transformations $ u = x^{2} - y^{2}$ and $v = 2xy$ , we need to plot the corresponding image in the $uv$-plane.
First we need to find $x$ and $y$ in terms of $u$ and $v$ ,
By completing the square , I got : $ (x^{2} + y^{2})^{2} = u^{2} + v^{2}$ 
=> $ x^{2} + y^{2} = \sqrt{u^{2} + v^{2}}$.
If I try to find equations for $x$ and $y$ from this , the situation becomes messy ..
Can anyone suggest a better way ?
Update : Using the equations $y = 2x$ and $y=4$ ,  I got the equations in $u$ and $v$ as : $v= (\dfrac{-4}{3}) u$ and $v^{2} = 64(u + 16)$ , and I was able to draw a graph as follows : 
 
(I know that doesn't look like a parabola , but still.. )
Now I am not able to find the image of the line $ y = 2x - 10$ , I tried solving $ u = x^{2} - y^{2}$ and $v = 2xy$ for the given $y$ but the equation ends up in terms of $x$ only.. Could anyone help ?
 A: Consider the substitution $x = r\cosh t, y = r\sinh t$.
Added:
$$u = x^2-y^2 =r^2\cosh^2 t - r^2\sinh^2t =r^2$$
$$v = 2xy =2 (r\cosh t)(r\sinh t) = r^2(2\cosh t\sinh t) = r^2\sinh 2t$$
So
$$ r = \sqrt u$$
$$ t = \frac 1 2 \sinh^{-1} \frac v u$$
and
$$ x = \sqrt u \cosh\left(\frac 1 2 \sinh^{-1} \frac v u\right)$$
$$ y = \sqrt u \sinh\left(\frac 1 2 \sinh^{-1} \frac v u\right)$$
Not very pretty, but it does express $x$ and $y$ in terms of $u$ and $v$.
A: Try building some intuition by watching what happens to the vertices and sides of the quadrilateral under the change-of-variables transformation $T(x,y) = (x^2-y^2, 2xy)$.
Example for an vertex:
$T(5,0) = (25,0)$
You should be able to easily find all the images of your vertices this way.  The only question that remains is what are the shapes of the curves that connect these vertices.
Example for a side: 
One side of the quadrilateral is defined by the curve $y=2x$ for $x \in [0,2]$.  You can think of this as a parametrized curve defined by the points$(x,2x)$ for $x \in [0,2]$.  Now watch what happens under the transformation:
$T(x,2x) = (\;(x)^2-(2x)^2, \;2(x)(2x)\;) = (-3x^2, 4x^2)$, still for $x \in [0,2]$.
When drawing a parametric curve, you are not obligated to move your pencil at a linear rate.  That is to say, you can re-parametrize the curve with $a=x^2$ without changing the curve itself.  Now you have $(-3a,4a)$ for $a \in [0,4]$.  Can you draw that?
