Minimizing Chord Length

The normal to the ellipse at the point $(a\cos\theta,b\sin\theta)$ has equation $ax\sin\theta - by\cos\theta \; = \; (a^2-b^2)\sin\theta\cos\theta$ which meets the ellipse again at the point $(a\cos\phi,b\sin\phi)$ , where $\begin{aligned} a^2\sin\theta\cos\phi - b^2\cos\theta\sin\phi & = \; (a^2-b^2)\sin\theta\cos\theta \\ \sin\big(\tfrac{\phi-\theta}{2}\big)\left[a^2\sin\theta\sin\big(\tfrac{\phi+\theta}{2}\big) + b^2\cos\theta\cos\big(\tfrac{\phi+\theta}{2}\big)\right] & = \; 0 \\ a^2\tan\theta\tan\big(\tfrac{\phi+\theta}{2}\big) + b^2 & = \; 0 \end{aligned}$ Thus the length of this chord of the ellipse is $D(\theta)$ , where $D(\theta)^2 \; = \; a^2(\cos\theta-\cos\phi)^2 + b^2(\sin\theta-\sin\phi)^2 \; = \; \frac{2a^2b^2\big(a^2 + b^2 - (a^2-b^2)\cos2\theta\big)^3}{\big(a^4 + b^4 - (a^4-b^4)\cos2\theta\big)^2}$ Turning points of this function, for $0 \le \theta \le \tfrac12\pi$ , occur at $\theta = 0,\tfrac12\pi$ . If $e > \tfrac{1}{\sqrt{2}}$ we also have a turning point at $\tfrac12\cos^{-1}\left(\frac{a^4 - 4a^2b^2+ b^4}{a^4-b^4}\right)$ (the condition $e > \tfrac{1}{\sqrt{2}}$ is what is required to ensure that $-1 < \tfrac{a^4-4a^2b^2+b^4}{a^4-b^4} < 1$ ) and $D(0)^2 \; = \; 4a^2 \hspace{1cm} D(\tfrac12\pi)^2 \; = \; 4b^2 \hspace{2cm} D\left(\tfrac12\cos^{-1}\left(\frac{a^4-4a^2b^2+b^4}{a^4-b^4}\right)\right) \; = \; \frac{27a^4b^4}{(a^2+b^2)^3} \; = \; 27a^2F\big(\tfrac{a}{b}\big)$ where $F(x) \; = \; \tfrac{x^4}{(1 + x^2)^3}$ . Now $F'(x) =\tfrac{2x^3(2-x^2)}{(1+x^2)^4}$ , and so $F(x) \le F(\sqrt{2}) = \tfrac{4}{27}$ . Thus the minimum chord length is $2b$ if $e < \tfrac{1}{\sqrt{2}}$ , and $\frac{3\sqrt{3}a^2b^2}{(a^2+b^2)^{\frac32}} \; = \; \boxed{a^2b^2 \left(\frac{3}{a^2+b^2}\right)^{\frac32}}$ if $e > \tfrac{1}{\sqrt{2}}$ .