Calculate Cotangent of Area Common to Two Ellipses

To find the area $\alpha$ of the common region lying inside both ellipses, we start by analyzing their equations: $E_1: x^2 + 4y^2 = 1 \implies y = \frac{1}{2}\sqrt{1 - x^2}$ $E_2: 4x^2 + y^2 = 1 \implies y = \sqrt{1 - 4x^2}$

Step 1: Find the Points of Intersection

Equating the two expressions for the ellipses: $x^2 + 4y^2 = 4x^2 + y^2 = 1 \implies 3x^2 = 3y^2 \implies x^2 = y^2$

Substituting $y^2 = x^2$ into $x^2 + 4y^2 = 1$: $5x^2 = 1 \implies x = \pm \frac{1}{\sqrt{5}}, \quad y = \pm \frac{1}{\sqrt{5}}$

Thus, in the first quadrant, the curves intersect at the point $P\left(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$.

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Step 2: Calculate the Area of the Common Region

Due to symmetry across both axes, the total area $\alpha$ is $4$ times the area of the region in the first quadrant ($A_1$).

In the first quadrant:

• For $0 \le x \le \frac{1}{\sqrt{5}}$, the upper boundary of the common region is given by $y = \frac{1}{2}\sqrt{1 - x^2}$.
• For $\frac{1}{\sqrt{5}} \le x \le \frac{1}{2}$, the upper boundary is given by $y = \sqrt{1 - 4x^2}$.

Therefore, the area in the first quadrant is: $A_1 = \int_{0}^{1/\sqrt{5}} \frac{1}{2}\sqrt{1 - x^2} \, dx + \int_{1/\sqrt{5}}^{1/2} \sqrt{1 - 4x^2} \, dx$

Evaluating the First Integral ($I_1$):

Let $x = \sin\theta \implies dx = \cos\theta \, d\theta$. $I_1 = \frac{1}{2} \int_{0}^{\arcsin(1/\sqrt{5})} \cos^2\theta \, d\theta = \frac{1}{4} \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\arcsin(1/\sqrt{5})}$

For $\theta_0 = \arcsin\left(\frac{1}{\sqrt{5}}\right)$, we have $\sin\theta_0 = \frac{1}{\sqrt{5}}$ and $\cos\theta_0 = \frac{2}{\sqrt{5}}$, so $\sin 2\theta_0 = 2\sin\theta_0\cos\theta_0 = \frac{4}{5}$.

$I_1 = \frac{1}{4} \left( \arcsin\left(\frac{1}{\sqrt{5}}\right) + \frac{2}{5} \right) = \frac{1}{4}\arcsin\left(\frac{1}{\sqrt{5}}\right) + \frac{1}{10}$

Evaluating the Second Integral ($I_2$):

Let $2x = \sin\phi \implies dx = \frac{1}{2}\cos\phi \, d\phi$. $I_2 = \frac{1}{2} \int_{\arcsin(2/\sqrt{5})}^{\pi/2} \cos^2\phi \, d\phi = \frac{1}{4} \left[ \phi + \frac{\sin 2\phi}{2} \right]_{\arcsin(2/\sqrt{5})}^{\pi/2}$

Since $\frac{\pi}{2} - \arcsin\left(\frac{2}{\sqrt{5}}\right) = \arccos\left(\frac{2}{\sqrt{5}}\right) = \arcsin\left(\frac{1}{\sqrt{5}}\right)$: $I_2 = \frac{1}{4} \left( \arcsin\left(\frac{1}{\sqrt{5}}\right) - \frac{2}{5} \right) = \frac{1}{4}\arcsin\left(\frac{1}{\sqrt{5}}\right) - \frac{1}{10}$

Total Area $\alpha$:

$A_1 = I_1 + I_2 = \frac{1}{2}\arcsin\left(\frac{1}{\sqrt{5}}\right)$ $\alpha = 4 A_1 = 2 \arcsin\left(\frac{1}{\sqrt{5}}\right)$

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Step 3: Compute $\cot \alpha$

Let $\theta = \arcsin\left(\frac{1}{\sqrt{5}}\right)$, so $\alpha = 2\theta$. From $\sin\theta = \frac{1}{\sqrt{5}}$, we get: $\tan\theta = \frac{1}{2}$

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

Therefore: $\cot\alpha = \frac{1}{\tan\alpha} = \frac{3}{4} = 0.75$