Two Twin Twirls

Ellipse $A$ Figure 1 As seen in figure 1, we place ellipse $A$ on a coordinate system. The major axis lies on ${x}'x$ its center is the origin. We set the radius of each semicircle equal to $1$ . In this setting, it is easy to see that the coordinates of point $P$ are $\left( -\sqrt{2},\dfrac{1}{\sqrt{2}} \right)$ .
Let the equation of ellipse $B$ be $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ . Then the equation of the tangent $l$ to the ellipse at $P$ is $\dfrac{{{x}_{P}}}{{{a}^{2}}}x+\dfrac{{{y}_{P}}}{{{b}^{2}}}x=1\Leftrightarrow \dfrac{-\sqrt{2}}{{{a}^{2}}}x+\dfrac{\frac{1}{\sqrt{2}}}{c}x=1\Leftrightarrow 2{{b}^{2}}x-{{a}^{2}}y+\sqrt{2}{{a}^{2}}{{b}^{2}}=0$ Hence, the slope of $l$ is ${{m}_{l}}=\dfrac{2{{b}^{2}}}{{{a}^{2}}} \ \ \ \ \ (1)$ At the same time, for the ellipse to have maximum eccentricity, $l$ must be tangent to the semicircle and in this case we have ${{m}_{l}}=\tan 45{}^\circ =1 \ \ \ \ \ (2)$

From $(1)$ and $(2)$ we get $\dfrac{2{{b}^{2}}}{{{a}^{2}}}=1\Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{1}{2}\Rightarrow 1-\dfrac{{{b}^{2}}}{{{a}^{2}}}=1-\dfrac{1}{2}\Rightarrow \sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}=\dfrac{1}{\sqrt{2}}\Rightarrow \boxed{e=\dfrac{1}{\sqrt{2}}}$

Ellipse $B$ Figure 2 Referring to figure 2, let the equation of ellipse $B$ be $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ .
Then, the radius of curvature at the ends of the major axis is $R=\dfrac{{{b}^{2}}}{a}$ In order for the yellow circles to be tangent to the ellipse, this value must be greater than or equal to the radius of the circles, which in turn is $\dfrac{a}{2}$ , i.e. $\dfrac{{{b}^{2}}}{a}\ge \dfrac{a}{2}$ .

Consequently, $\dfrac{{{b}^{2}}}{{{a}^{2}}}\ge \dfrac{1}{2}\Rightarrow 1-\dfrac{{{b}^{2}}}{{{a}^{2}}}\le 1-\dfrac{1}{2}\Rightarrow \sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}\le \dfrac{1}{\sqrt{2}}\Rightarrow e\le \dfrac{1}{\sqrt{2}}$ Since the ellipse has maximum eccentricity, it holds $\boxed{e = \dfrac{1}{\sqrt{2}}}$

It is clear that both ellipse $A$ and ellipse $B$ have the same eccentricity.