Question

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 $\tan(\alpha/2)$ is {

Hint:

Using polar coordinates for intersection area problems involving central conics (ellipses/hyperbolas centered at the origin) often transforms a difficult Cartesian integral into a simple trigonometric one.

Answer 1

Correct Answer: 2

Step 1: Understanding the Question:
The area $\alpha$ is the intersection area of two ellipses. By symmetry, the area is 8 times the area of a portion in the first quadrant bounded by $y=x$ and one of the ellipses.

Step 2: Key Formula or Approach:

• Area in polar coordinates: $\alpha = \int \frac{1}{2} r^2 d\theta$.

• Region consists of 8 equal sectors from $\theta = 0$ to $\pi/4$.

Step 3: Detailed Explanation:

• In polar form, $x^2 + 4y^2 = 1$ becomes $r^2(\cos^2 \theta + 4\sin^2 \theta) = 1 \implies r^2 = \frac{1}{1 + 3\sin^2 \theta}$.

• Area $\alpha = 8 \times \int_0^{\pi/4} \frac{1}{2} r^2 d\theta = 4 \int_0^{\pi/4} \frac{1}{1 + 3\sin^2 \theta} d\theta$.

• To integrate, divide numerator and denominator by $\cos^2 \theta$:
$\alpha = 4 \int_0^{\pi/4} \frac{\sec^2 \theta}{1 + \tan^2 \theta + 3\tan^2 \theta} d\theta = 4 \int_0^{\pi/4} \frac{\sec^2 \theta}{1 + 4\tan^2 \theta} d\theta$.

• Let $u = \tan \theta, du = \sec^2 \theta d\theta$. Limits: $0 \to 1$.
$\alpha = 4 \int_0^1 \frac{du}{1 + 4u^2} = 4 \left[ \frac{1}{2} \tan^{-1}(2u) \right]_0^1 = 2 \tan^{-1}(2)$.

• Thus $\alpha = 2 \tan^{-1}(2) \implies \alpha/2 = \tan^{-1}(2)$.

• $\tan(\alpha/2) = \tan(\tan^{-1}(2)) = 2$.

Step 4: Final Answer:
The value of $\tan(\alpha/2)$ is 2.