JEE Advanced 2026 · Paper 2, Mathematics question 18

By symmetry the common region is 8 copies of its part with $0\le\theta\le\frac\pi4$, where the binding curve is $4x^2+y^2=1$ (it reaches only $x=\frac12$ on the $x$-axis). In polar form $r^2=\dfrac1{4\cos^2\theta+\sin^2\theta}$:
$$\alpha=8\int_0^{\pi/4}\frac{r^2}2\,d\theta=4\int_0^{\pi/4}\frac{\sec^2\theta\,d\theta}{4+\tan^2\theta}=4\cdot\frac12\tan^{-1}\frac12=2\tan^{-1}\frac12.$$
So $\tan\alpha=\dfrac{2\cdot\frac12}{1-\frac14}=\dfrac43$ and $\cot\alpha=\dfrac34=0.75$.