Based of the half angle of a double cone, and the angle subtended by a plane with the cone's axis, is it possible to determine the eccentricity of the conic section which is formed by this.
I think it should be
$$\tan{π((π/2)-b)\over 2a},$$
where
$b=\ $angle subtended by plane with axis of cone
and
$a=\ $half angle of the cone.
In the case of an ellipse, you can find all the relevant formulae in this old answer of mine (with an appendix here).
We can in particular compute the ratio $\eta={\text{minor axis}\over\text{major axis}}$, which translated into your notations becomes: $$ \eta^2={\sin(b-a)\sin(b+a)\over\cos^2 a}. $$ The square of the eccentricity can then be found as $e^2=1-\eta^2$, that is $$ e^2=1-{\sin(b-a)\sin(b+a)\over\cos^2 a}={\cos^2b\over\cos^2a}, $$ where I used some standard trig equalities in the last passage. Hence we have the nice result: $$ e={\cos b\over\cos a}. $$
As it often happens, this formula gives the correct result even in the case of a hyperbola or parabola.