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

The given ellipses are $E_1: x^2 + 4y^2 = 1$ and $E_2: 4x^2 + y^2 = 1$.

By symmetry, the common region is symmetric about both the coordinate axes and the lines $y = x$ and $y = -x$. The total area $\alpha$ is $8$ times the area of the region in the first quadrant bounded by $\theta = 0$ and $\theta = \dfrac{\pi}{4}$.

In polar coordinates, substituting $x = r\cos\theta$ and $y = r\sin\theta$ into the equation of $E_2$ (which is the inner boundary for $0 \le \theta \le \dfrac{\pi}{4}$), we get:
$4r^2\cos^2\theta + r^2\sin^2\theta = 1 \Rightarrow r^2 = \dfrac{1}{4\cos^2\theta + \sin^2\theta}$

The area of this sector is given by:
$A = \dfrac{1}{2} \int_{0}^{\dfrac{\pi}{4}} r^2 d\theta = \dfrac{1}{2} \int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{4\cos^2\theta + \sin^2\theta} d\theta$

Dividing the numerator and the denominator by $\cos^2\theta$:
$A = \dfrac{1}{2} \int_{0}^{\dfrac{\pi}{4}} \dfrac{\sec^2\theta}{4 + \tan^2\theta} d\theta$

Substituting $t = \tan\theta$, we have $dt = \sec^2\theta d\theta$. The limits change from $0$ to $1$:
$A = \dfrac{1}{2} \int_{0}^{1} \dfrac{dt}{4 + t^2} = \dfrac{1}{2} \left[ \dfrac{1}{2} \arctan\left(\dfrac{t}{2}\right) \right]_{0}^{1} = \dfrac{1}{4} \arctan\left(\dfrac{1}{2}\right)$

The total area $\alpha$ of the common region is:
$\alpha = 8A = 8 \times \dfrac{1}{4} \arctan\left(\dfrac{1}{2}\right) = 2 \arctan\left(\dfrac{1}{2}\right)$

We need to find the value of $\cot\alpha$:
$\cot\alpha = \cot\left(2 \arctan\left(\dfrac{1}{2}\right)\right)$

Let $\phi = \arctan\left(\dfrac{1}{2}\right)$, which implies $\tan\phi = \dfrac{1}{2}$.

Using the double angle formula for tangent:
$\tan(2\phi) = \dfrac{2\tan\phi}{1 - \tan^2\phi} = \dfrac{2 \left(\dfrac{1}{2}\right)}{1 - \left(\dfrac{1}{2}\right)^2} = \dfrac{1}{1 - \dfrac{1}{4}} = \dfrac{4}{3}$

Therefore, $\cot\alpha = \dfrac{1}{\tan(2\phi)} = \dfrac{3}{4}$.

Answer: $3/4$