JEE Main 2025 Jan 22 Shift 2, Mathematics Q20: Area as Definite Integral

Rewrite both curves with respect to $X=x-2$, $Y=y$:

$y=(x-2{)}^2\Rightarrow Y=X^2$, an upward parabola through the origin in $X,Y$ coordinates.

$y^2=16-8x=-8(x-2)\Rightarrow Y^2=-8X$, a left-opening parabola also through the origin.

So in the shifted frame we need the area enclosed between $Y=X^2$ and $Y^2=-8X$. These two parabolas pass through the origin; finding the other intersection:

From $Y=X^2$, substitute into $Y^2=-8X$: $X^4=-8X\Rightarrow X(X^3+8)=0\Rightarrow X=0$ or $X=-2$ (real).

At $X=-2$: $Y=4$. So intersection points are $(0,0)$ and $(-2,4)$.

Using the standard result for the area between $y=x^2/(4a)$-type parabolas opening orthogonally, the enclosed area between $Y^2=4aX$ (here $4a=8$, so $a=2$, but with $X$ negative, treat $|X|$) and $X^2=4bY$ (here $4b=1$, so $b=1/4$) is $\frac{16ab}{3}$.

With $a=2$ and $b=1/4$: area $=\frac{16\cdot 2\cdot \frac{1}{4}}{3}=\frac{8}{3}$.