Equable ellipses II

Say the semi-major axis is $a$ , the semi-minor axis $b$ , and the eccentricity is $e$ , where $e=\sqrt{1-\frac{b^2}{a^2}}$

A parametric equation of this ellipse is $x(t)=a\cos t,\;\;\;y(t)=b\sin t$

We can find the perimeter of the ellipse, $c$ , as

$\begin{aligned} c&=4\int_0^{\frac{\pi}{2}} \sqrt{\dot{x}^2+\dot{y}^2}\; dt \\ &=4\int_0^{\frac{\pi}{2}} \sqrt{a^2\sin^2 t+b^2 \cos^2 t}\; dt \\ &=4\int_0^{\frac{\pi}{2}} \sqrt{a^2\sin^2 t+a^2 \left(1-e^2 \right) \cos^2 t}\; dt \\ &=4a\int_0^{\frac{\pi}{2}} \sqrt{\sin^2 t+\left(1-e^2 \right) \cos^2 t}\; dt \\ &=4a\int_0^{\frac{\pi}{2}} \sqrt{1-e^2 \cos^2 t}\; dt \end{aligned}$

The eccentricity of an ellipse lies in the range $0\le e<1$ . For a fixed $a$ , it's clear that the perimeter decreases as $e$ increases.

In the limit $e \nearrow 1$ , $c \searrow 4a\int_0^{\frac{\pi}{2}} \sqrt{1- \cos^2 t}\; dt = 4a$

In an equable ellipse, $\pi ab=c$ ; so $\begin{aligned} \pi ab &> 4a \\ b &>\frac{4}{\pi} \approx \boxed{1.2732} \end{aligned}$

As to the bonus, no; it's not really surprising if you consider that there is also a minimum width for an equable rectangle.