(1) If the area of the ellipse equals the area of the annulus then $\pi \mathrm{ab}=\pi b^2-\pi a^2$ and so $\mathrm{ab}=b^2-a^2$. Then, dividing by $a^2$,

${\displaystyle b/a=(b/a{)}^2-1.}$

The ratio we want to find is $b/a$, the ratio of the longer to the shorter axis of the ellipse. So let $b/a=x$ then

${\displaystyle x^2-x-1=0.}$

Using quadratic formula:

${\displaystyle x=\frac{1\pm \sqrt{5}}{2}}$

We choose the positive root knowing that the ratio $b/a$ is positive so this ratio is equal to the golden ratio.

(2) Note that $R$ appears itself in the nested root. Therefore we can say

${\displaystyle R=\sqrt{(}1+R)}$

and so

${\displaystyle R^2-R-1=0.}$

We have a quadratic of the same form as above. Hence we find $R$ to be the golden ratio.