Area, JEE, 2026

#Calculus#AreasBetweenTwoCurves#PolarCoordinates

Problem Statement

Consider the ellipses given by

$$ 4x^2+y^2 =1 \ \text{and} \ x^2+4y^2 $$

If $\alpha$ is the area of the common regions that lies inside both the given ellipses, then the value of $\cot{\alpha}$ is

Let us begin by plotting the curves

Since the two curves can be obtained by interchanging $x$ and $y$, therefore the curves are simply reflections of each other in the line $y=x.$ The symmetry guarantees that the area of the region being asked is $8$ times the area of the region $R_{1}.$ It would be a terrible idea to set up the integral in a Cartesian system. Given the symmetry of the ellipse, let us invoke polar co-ordinates. Let the radius from the origin be of length $\rho$ and say that the radius vector makes an angle $\theta$ with the positive $x-$ axis. The co-ordinates of the point are $(\rho \cos{\theta}, \rho \sin{\theta}).$ Since it lies on the ellipse $4x^2+y^2=1. $ Therefore, we have

$$ 4 \rho^2 \cos^2{\theta} + \rho^2 \sin^2{\theta} =1 \implies \rho^2 = \frac{1}{4 \cos^2{\theta}+ \sin^2{\theta} }. $$

Since the boundaries of the region $R_{1}$ are contained by the rays $ \theta =0 $ and $\theta = \frac{\pi}{4}.$ So the given area is

$$ \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \rho^2(\theta ) d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{4 \cos^2{\theta}+ \sin^2{\theta} } d \theta $$

$$ \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{\tan^2\theta+4 } \cdot \sec^2 \theta d \theta $$

Put $ \tan \theta = t \implies \sec^2 \theta d\theta = dt. $ The given integral transforms into

$$ \frac{1}{2} \int_{0}^{1} \frac{1}{t^2+4} dt = \frac{1}{4} \left[\tan^{-1}\left(\frac{t}{2} \right )\right]_{0}^{1} = \frac{1}{4} \tan^{-1}\left (\frac{1}{2} \right) $$

$$ \implies \alpha = 8 \cdot \frac{1}{4} \tan^{-1}\left (\frac{1}{2} \right) = 2 \tan^{-1}\left (\frac{1}{2} \right) $$

$$ \implies \cot{\left( 2 \tan^{-1}\left (\frac{1}{2} \right) \right)} =\frac{3}{4}. $$

Bibliography & References

[1]
JEE 2026