JEE Advanced 2026 Paper 2, Mathematics Section 4 Q4: Area as Definite Integral

The two ellipses are reflections of each other across the line $y=x$, and their boundary in the first quadrant meets at $P=\left(\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)$. By symmetry, the common region has eightfold symmetry about the axes and the lines $y=\pm x$.

Consider the slice in the first quadrant between $y=x$ and the upper boundary of ellipse 1 ($y=\frac{\sqrt{1-x^2}}{2}$), from $x=0$ to $x=\frac{1}{\sqrt{5}}$. Call this area $A$. Then the total common area is $\alpha =8A$.

$A={\int }_0^{1/\sqrt{5}}\left(\frac{\sqrt{1-x^2}}{2}-x\right)dx$.

Using $\int \sqrt{1-x^2}dx=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x$:

$\alpha =8A=8{\left[\frac{x\sqrt{1-x^2}}{4}+\frac{{\sin }^{-1}x}{4}-\frac{x^2}{2}\right]}_0^{1/\sqrt{5}}$.

At $x=\frac{1}{\sqrt{5}}$: $\sqrt{1-x^2}=\frac{2}{\sqrt{5}}$, so the bracket equals $\frac{1}{4\sqrt{5}}\cdot \frac{2}{\sqrt{5}}+\frac{1}{4}{\sin }^{-1}\frac{1}{\sqrt{5}}-\frac{1}{10}=\frac{1}{10}+\frac{1}{4}{\sin }^{-1}\frac{1}{\sqrt{5}}-\frac{1}{10}=\frac{1}{4}{\sin }^{-1}\frac{1}{\sqrt{5}}$.

So $\alpha =8\cdot \frac{1}{4}{\sin }^{-1}\frac{1}{\sqrt{5}}=2{\sin }^{-1}\frac{1}{\sqrt{5}}$.

Note that ${\sin }^{-1}\frac{1}{\sqrt{5}}={\tan }^{-1}\frac{1}{2}$ (right triangle with opposite 1, adjacent 2, hypotenuse $\sqrt{5}$).

So $\alpha =2{\tan }^{-1}\frac{1}{2}$. Using $\tan (2\theta )=\frac{2\tan \theta }{1-{\tan }^2\theta }$ with $\tan \theta =\frac{1}{2}$: $\tan \alpha =\frac{1}{1-1/4}=\frac{4}{3}$.

$\therefore \cot \alpha =\frac{3}{4}=0.75$.