Football in Rugby Ball?

$\dfrac{{x}^2}{{(4a)}^2}+\dfrac{{y}^2}{{(2a)}^2}=1$

Let $(4a\cos\theta,2a\sin\theta)$ be a point on the ellipse.

Equation of normal to the ellipse at $\theta$ is given by

$\dfrac{4ax}{\cos\theta}-\dfrac{2ay}{\sin\theta}={(4a)}^2-{(2a)}^2\\\Rightarrow \dfrac{4ax}{\cos\theta}-\dfrac{2ay}{\sin\theta}={12a}^2$

On substituting the point $(a,0)$ in the equation obtained, we get

$\cos\theta=\dfrac{1}{3}\\\Rightarrow \sin\theta=\dfrac{2\sqrt{2}}{3}$

So the point where the inscribed circle touches the ellipse is $\left(\dfrac{4a}{3},\dfrac{4\sqrt{2}}{3}\right)$ . Also the distance of this point from $(a,0)$ give us the radius of the inscribed circle.

$\begin{aligned}r&=\sqrt{\left(\dfrac{4a}{3}-a\right)^2+\left(\dfrac{4\sqrt{2}}{3}-0\right)^2}\\&\huge\color{#3D99F6}{=\boxed{\sqrt{\dfrac{11}{3}}a}}\end{aligned}$

Any point on the ellipse is $\large{(4a \cos t, 2a \sin t)}$ .

We have to maximize the distance between $(a,0)$ and $(4a \cos t, 2a \sin t)$ .For this $t$ should be minimum because as $t \to 0$ the coordinates tend toward the major axis , so our distance will be maximum.

Now $\large{R^2=(4a \cos t -a)^2+4a^2 \sin^2 t}$

$\large{\Rightarrow R^2 =a^2+16a^2 \cos^2 t -8a^2 \cos t+4a^2 \sin^2 t}$

let $R^2 =A$

Taking its derivative we get $\large{\frac{d A}{d t}=\sin t \left(8a^2-24 a^2 \cos t \right)}$

Equating to zero we get $\sin t=0$ or $\cos t=\frac{1}{3}$ , by the second derivative test we find that $t$ is minimized for $\cos t=\frac{1}{3}$

Thus , $\large{R^2=a^2 \left( (\frac{4}{3}-1)^2+4 \left(\frac{\sqrt{8}}{3} \right)^2 \right)}$

$\large{R=\sqrt{\frac{11}{3}}a}$