Question 18

Step 1: Express the ellipses in polar coordinates.

$E_1: x^2 + 4y^2 = 1 \implies r^2(\cos^2 \theta + 4 \sin^2 \theta) = 1 \implies r_1^2 = \frac{1}{1 + 3\sin^2 \theta}$.

$E_2: 4x^2 + y^2 = 1 \implies r^2(4\cos^2 \theta + \sin^2 \theta) = 1 \implies r_2^2 = \frac{1}{4 - 3\sin^2 \theta}$.

Step 2: Determine the intersection and the inner boundary.

The ellipses intersect at $\theta = \frac{\pi}{4}$. By symmetry, the total area $\alpha$ is 8 times the area in the sector $[0, \pi/4]$. In $[0, \pi/4]$, $r_2 < r_1$ because at $\theta=0$, $r_2^2 = 1/4$ and $r_1^2 = 1$.

Step 3: Calculate the area $\alpha$.

$\alpha = 8 \times \int_0^{\pi/4} \frac{1}{2} r_2^2 d\theta = 4 \int_0^{\pi/4} \frac{d\theta}{4 - 3\sin^2 \theta}$.

Divide numerator and denominator by $\cos^2 \theta$:

$\alpha = 4 \int_0^{\pi/4} \frac{\sec^2 \theta d\theta}{4\sec^2 \theta - 3\tan^2 \theta} = 4 \int_0^{\pi/4} \frac{\sec^2 \theta d\theta}{4(1 + \tan^2 \theta) - 3\tan^2 \theta} = 4 \int_0^{\pi/4} \frac{\sec^2 \theta d\theta}{4 + \tan^2 \theta}$.

Let $t = \tan \theta, dt = \sec^2 \theta d\theta$:

$\alpha = 4 \int_0^1 \frac{dt}{4 + t^2} = 4 \left[ \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \right]_0^1 = 2 \tan^{-1}\left(\frac{1}{2}\right)$.

Step 4: Find $\cot \alpha$.

Let $\beta = \tan^{-1}(1/2)$, so $\alpha = 2\beta$.

$\tan \alpha = \tan(2\beta) = \frac{2\tan\beta}{1 - \tan^2\beta} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{3/4} = \frac{4}{3}$. Therefore, $\cot \alpha = \frac{3}{4} = 0.75$.