Some people insist that to create a circle kissing the vertex of an ellipse or a hyperbola we need to calculate the second derivative, but I think it can be done geometrically, I came up with this method using GeoGebra, I need proof to show that this is the same result that we will get from the second derivative:
Let $P$ be an ellipse/hyperbolus with center $S$, foci $F,F'$, principal vertices $A,A'$, and kissing circles $P$ at $A'$, centers $O$, then:
$\frac{SO}{SF}=\frac{SF}{SA}$
$\frac{SO}{SF}=\frac{SF}{SA}$
$\frac{SO}{SF}=\frac{SF}{SA}$
There are several ways to construct the osculating circle at a generic point of a conic, see for instance this old answer of mine. Translating that result for the case of the ellipse in this question, we obtain: $$ {2\over AO}={1\over AF}+{1\over AF'}. $$ Substituting there $AO=SA-SO$, $AF=SA-SF$, $AF'=SA+SF$ we get: $$ {1\over SA-SO}={SA\over SA^2-SF^2}, $$ which is equivalent to your result.
An analogous computation can be done for the hyperbola.
Another geometrical way to construct the osculating circle of an ellipse can be found for instance in Drew's book, Prop. XXVI and XXVII.