<s>[INST] Given $c =\sqrt {a^2-b^2}$, $c= ea$, conclude that $\frac {2\pi ab}{(a+c)T} =\frac {2\pi a}T\sqrt {\frac {1-e}{1+e}}$ Given $c =\sqrt {a^2-b^2}$, $c= ea$, conclude that $$\frac {2\pi ab}{(a+c)T} =\frac {2\pi a}T\sqrt {\frac 
{1-e}{1+e}}$$
Sorry for the awful formatting of this question. I know this is a pretty simple problem, but for some reason I cannot figure it out. The furthest I have gotten is reducing the b in the numerator to $\sqrt {a^2(1-e)}$, but then I get stuck. I appreciate any help I can get. Thanks in advance.
 [/INST] Since $\;c =\sqrt {a^2-b^2},\;$ we get $\;b=\sqrt {a^2-c^2}$.
Moreover,
$$\frac {2\pi ab}{(a+c)T} =\frac{2\pi a\sqrt{a^2-c^2}}{T(a+c)}=\frac{2\pi a}T\sqrt{\frac 
{(a+c)(a-c)}{(a+c)^2}}=\frac{2\pi a}T\sqrt{\frac 
{a-c}{a+c}}=\frac{2\pi a}T\sqrt{\frac 
{a-ea}{a+ea}}=\frac{2\pi a}T\sqrt{\frac 
{a(1-e)}{a(1+e)}}=\frac{2\pi a}T\sqrt{\frac 
{1-e}{1+e}}.$$
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