Arc length inequality on ellipse

Question

Suppose $E$ is the ellipse $(\frac{x}{a})^2+(\frac{y}{b})^2=1.$ For $P,Q\in E$ let $\sigma(P,Q)$ denote the length of the minor arc connecting $P$ to $Q$. Find the maximum value for $A>0$ such that $|P-Q|\geq A \sigma(P,Q)$, for all $P,Q$ on $E.$,

Answer 1

Some ideas on the problem. Let $\alpha : \mathbf R\to\mathbf R^2$, $\alpha(t)=(a\cos(t),b\cos(t))$ be the desired ellipse. Its arclength from $Q=\alpha(q)$ to $P=\alpha(p)$ is then $$\ell(P)=\int_q^p |\alpha'(t)|\,dt=\int_q^p\sqrt{a^2\sin^2(t)+b^2\cos^2(t)}\,dt.$$ By the mean value theorem, there is $0\leq\xi\leq p-q$ such that $$\ell(P, Q)=(p-q)\sqrt{a^2\sin^2(q+\xi)+b^2\cos^2(q+\xi)}.$$ Keep in mind $$|P-Q|=|a\cos(p)-a\cos(q),b\sin(p)-b\sin(q)|\\=\sqrt{a^2(\cos(p)-\cos(q))^2+b^2(\sin(p)-\sin(q))^2}.$$

Now, divide $\frac{|P-Q|}{\ell(P,Q)}$ and try to bound it from below. For example, $$|P-Q|\geq\sqrt{2ab|(\cos(p)-\cos(q))(\sin(p)-\sin(q))|}$$ and (since $|\int f|\leq(\int f^2)^{1/2}$) $$\ell\leq \sqrt{\int_q^pa^2\sin^2(t)+b^2\cos^2(t)\,dt}.$$ Intuitively, $A$ should be obtained comparing the distance and arclength between $(-a,0)$ and $(a,0)$ if $a>b$. I'm still thinking about the problem.