If the area of a square circumscribing an ellipse is A square units. The maximum distance of a normal from the centre of ellipse is r, then the value of  $\left[\frac{A}{14\cdot r^2}\right]$ is (Where   $[\cdot ]$ denotes greatest integer function) (Given eccentricity of ellipse is 0.5)

PQRS is a square

$$\begin{gathered} \Rightarrow {\mathcal{l}}^2=A\Rightarrow \mathcal{l}=\sqrt{A}\text{ where }\sqrt{2}\mathcal{l}=2\sqrt{a^2+b^2} \\ \Rightarrow \sqrt{A}=\sqrt{2(a^2+b^2)}\Rightarrow a^2+b^2=\frac{A}{2}\text{ and }a-b=r \\ \Rightarrow a^2+b^2-2ab=r^2\Rightarrow 2ab=(\frac{A}{2}-r^2) \\ {(a+b)}^2=\frac{A}{2}+(\frac{A}{2}-r^2)=(A-r^2)\Rightarrow a+b=\sqrt{A-r^2} \\ \Rightarrow 2a=r+\sqrt{A-r^2},2b=\sqrt{A-r^2}-r \end{gathered}$$

$e^2=1-\left(\frac{b^2}{a^2}\right)\Rightarrow \frac{1}{4}=1-{\left(\frac{\sqrt{A-r^2}-r}{\sqrt{A-r^2+r}}\right)}^2\Rightarrow \frac{r^2{(\sqrt{\frac{A}{r^2}-1}-1)}^2}{r^2{(\sqrt{\frac{A}{r^2}-1}+1)}^2}=\frac{3}{4}\Rightarrow \frac{\sqrt{\frac{A}{r^2}-1}+1}{\sqrt{\frac{A}{r^2}-1}-1}=\frac{2}{\sqrt{3}}$

$\sqrt{3}\sqrt{\frac{A}{r^2}-1}+\sqrt{3}=2\sqrt{\frac{A}{r^2}-1}-2;(2-\sqrt{3})\sqrt{\frac{A}{r^2}-1}=(2+\sqrt{3})\Rightarrow \frac{A}{r^2}-1={(2+\sqrt{3})}^4$

$\therefore \left[\frac{A}{r^2}\right]=194$