JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Area Under Curves

The parabola $y^2=9x$ and the line $y=3x-6$ meet where $\frac{y^2}{9}=\frac{y+6}{3}$, i.e. $y^2-3y-18=0$, at $y=6(x=4)$ and $y=-3(x=1)$. The bounded region with $0\le 9x\le y^2$ and $y\ge 3x-6$ lies to the right of the $y$-axis, left of the parabola and above the line, closing off between $y=-6$ and $y=0$. For $-3\le y\le 0$ its right edge is the parabola $x=\frac{y^2}{9}$, and for $-6\le y\le -3$ its right edge is the line $x=\frac{y+6}{3}$, so $A={\int }_{-3}^0\frac{y^2}{9}dy+{\int }_{-6}^{-3}\frac{y+6}{3}dy=1+\frac{3}{2}=\frac{5}{2}$. Therefore $6A=15$.