How would I calculate the inverse of
$F(x,y)=\left( \textrm{arctan} \left(\frac{ay}{x}\right),\frac{x^2+a^2y^2}{2a}\right)$
$a$ is a constant.
How would I calculate the inverse of
$F(x,y)=\left( \textrm{arctan} \left(\frac{ay}{x}\right),\frac{x^2+a^2y^2}{2a}\right)$
$a$ is a constant.
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First set $F(x,y) = (X,Y)$
Then you have $$ay = x \textrm{tan}X$$ and $$x^2 + (a y )^2 = 2 a Y$$ Sub the first eqn into the second to get $$x^2 = 2a Y /(1 + (\textrm{tan} X)^2 )$$
This should get you started.