A geometry problem by HCR

With the parametrisation $x = a\cos\theta$ , $y = b\sin\theta$ , we have $\frac{ds}{d\theta} \; = \; \sqrt{a^2\sin^2\theta + b^2\cos^2\theta} \; = \; a\sqrt{1 - e^2\cos^2\theta}$ so that the perimeter of the ellipse is $P \; = \; a\int_0^{2\pi} \sqrt{1 - e^2\cos^2\theta}\,d\theta \; = \; 4aE(e^2)$ where $E$ is the complete elliptic integral of the second kind. Thus the perimeter of the ellipse is $2\pi k a$ where $k \; = \; \frac{2}{\pi}E(e^2)$ which is equal to $\boxed{0.81255}$ when $e = 0.8$ .