The minimum length of focal chord $AB$ of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ passing through $A$ is___ 
$\left(\mathrm{Given}A=\left(\frac{8}{5},\frac{3\sqrt{3}}{5}\right)\right)$

$\therefore 2\cos {\theta }_1=\frac{8}{5}\Rightarrow \cos {\theta }_1=\frac{4}{5}\Rightarrow \tan \frac{{\theta }_1}{2}=\sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}=\frac{1}{3}\text{ and }\mathrm{e}=\frac{1}{2}$
$\begin{array}{l}\therefore \tan \frac{{\theta }_2}{2}\times \frac{1}{3}=\frac{-\frac{1}{2}}{\frac{3}{2}}\\ \frac{{\theta }_2}{2}=\frac{3\pi }{4}\Rightarrow {\theta }_2=\frac{3\pi }{2}\\ \therefore B=(0,-\sqrt{3})\Rightarrow AB=2+2-\frac{1}{2}\times \frac{8}{5}=4-\frac{4}{5}=\frac{16}{5}\end{array}$