Consider the ellipses given by $$x^2+4y^2=1$$ and $$4x^2+y^2=1$$.

If $$\alpha$$ is the area of the common region that lies inside both the given ellipses, then the value of $$\cot\alpha$$ is ___.

$$A = \int_{0}^{\frac{1}{\sqrt{5}}} \left( \frac{\sqrt{1-x^2}}{2} - x \right) dx$$

$$\alpha = 8A = 8 \left[ \int_{0}^{\frac{1}{\sqrt{5}}} \left( \frac{\sqrt{1-x^2}}{2} - x \right) dx \right]$$

$$\alpha = 8 \left[ \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1}x - \frac{x^2}{2} \right]_{0}^{\frac{1}{\sqrt{5}}}$$

$$\alpha = 8 \left[ \frac{1}{4\sqrt{5}}\sqrt{\frac{4}{5}} + \frac{1}{4}\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) - \frac{1}{10} \right]$$

$$\alpha = 2\sin^{-1}\frac{1}{\sqrt{5}}$$

$$\cot\alpha = \cot\left( 2\sin^{-1}\frac{1}{\sqrt{5}} \right)$$

$$= \cot\left( 2\tan^{-1}\frac{1}{2} \right)$$

$$= \cot\left( \tan^{-1}\left( \frac{2 \times \frac{1}{2}}{1 - \frac{1}{4}} \right) \right)$$

$$= \cot\left( \tan^{-1}\left( \frac{4}{3} \right) \right)$$ $$ = 0.75$$