JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Area Under Curves

The curves $y^2=ax$ and $x^2=ay$ meet at the origin and at $P(a,a)$, and the region between them has area ${\int }_0^a(\sqrt{ax}-\frac{x^2}{a})dx=\frac{a^2}{3}$. The chord $OP$ is the line $y=x$, so where $x=b$ meets it the point $Q$ is $(b,b)$ and the triangle $OQR$ has area $\frac{1}{2}b\cdot b=\frac{1}{2}$, giving $b=1$. Since $x=b$ bisects the region, the part up to $x=1$ is half the total: ${\int }_0^1(\sqrt{ax}-\frac{x^2}{a})dx=\frac{2}{3}\sqrt{a}-\frac{1}{3a}=\frac{a^2}{6}$. Clearing denominators gives $4a\sqrt{a}-2=a^3$, and squaring $a^3+2=4a^{3/2}$ leads to $a^6-12a^3+4=0$.