Consider the ellipses given by x² + 4y² = 1 and 4x² + y² = 1. If α is the area of the common region enclosed by both the ellipses, then the value of tan(α/2) is _______.

The correct answer is tan(α/2) = 2.

We are given two ellipses:

E₁: x² + 4y² = 1  (semi-axes: a = 1 along x, b = 1/2 along y)
E₂: 4x² + y² = 1  (semi-axes: a = 1/2 along x, b = 1 along y)

Note that E₂ is simply E₁ rotated 90°, so both ellipses have equal areas. The figure has four-fold symmetry about both coordinate axes, and also symmetry about the lines y = x and y = -x.

Step 1 — Find the intersection points.

Subtract the two equations:

(x^2+4y^2)-(4x^2+y^2)=0\Rightarrow 3y^2-3x^2=0\Rightarrow y=\pm x

Substitute y = x into E₁:

x^2+4x^2=1\Rightarrow 5x^2=1\Rightarrow x=\pm \frac{1}{\sqrt{5}}

The four intersection points are $(\pm \frac{1}{\sqrt{5}},\pm \frac{1}{\sqrt{5}})$.

Step 2 — Area of each ellipse.

For an ellipse with semi-axes a and b, the area is πab. For both E₁ and E₂:

A_{E1}=A_{E2}=\pi ·1·\frac{1}{2}=\frac{\pi }{2}

Step 3 — Compute the overlapping (intersection) area using polar coordinates.

Converting to polar coordinates (x = r cosθ, y = r sinθ), the two ellipses become:

r_1=\frac{1}{\sqrt{{\cos }^2\theta +4{\sin }^2\theta }},r_2=\frac{1}{\sqrt{4{\cos }^2\theta +{\sin }^2\theta }}

By comparing denominators: r₂ < r₁ when 4cos²θ + sin²θ > cos²θ + 4sin²θ, i.e., when tanθ < 1, i.e., for 0 ≤ θ < π/4. Similarly, r₁ < r₂ for π/4 < θ ≤ π/2.

The first-quadrant area of the intersection is:

A_{\text{int, Q1}}=\frac{1}{2}{\int }_0^{\pi /4}{r}_2^2\mathrm{d\theta }+\frac{1}{2}{\int }_{\pi /4}^{\pi /2}{r}_1^2\mathrm{d\theta }

By the substitution θ → π/2 - θ, both integrals are equal, so:

A_{\text{int, Q1}}={\int }_0^{\pi /4}\frac{\mathrm{d\theta }}{4{\cos }^2\theta +{\sin }^2\theta }

Dividing numerator and denominator by cos²θ and using the substitution t = tanθ, dt = sec²θ dθ:

={\int }_0^{\pi /4}\frac{{\sec }^2\theta \mathrm{d\theta }}{4+{\tan }^2\theta }={\int }_0^1\frac{\mathrm{dt}}{4+t^2}={\left[\frac{1}{2}\arctan \frac{t}{2}\right]}^1{}_0=\frac{1}{2}\arctan \frac{1}{2}

The total intersection area (across all 4 quadrants) is:

A_{\text{intersection}}=4\times \frac{1}{2}\arctan \frac{1}{2}=2\arctan \frac{1}{2}

Step 4 — Find α, the total area enclosed by both ellipses (union).

By the inclusion-exclusion principle:

\alpha =A_{E1}+A_{E2}-A_{\text{intersection}}=\frac{\pi }{2}+\frac{\pi }{2}-2\arctan \frac{1}{2}=\pi -2\arctan \frac{1}{2}

Step 5 — Compute tan(α/2).

Dividing α by 2:

\frac{\alpha }{2}=\frac{\pi }{2}-\arctan \frac{1}{2}

Let φ = arctan(1/2), so that tan φ = 1/2. Then:

\tan \frac{\alpha }{2}=\tan (\frac{\pi }{2}-\phi )=\cot \phi =\frac{1}{\tan \phi }=\frac{1}{1/2}=2

Therefore, tan(α/2) = 2.