Animated Solution for Mathematics - Definite Integration: Let $A_1=\{(x,y):|x|\le y^2,|x|+2y\le 8\}$ and $A_2=\{(x,y):|x|+|y|\le k\}.$ If $27\text{ (Area }{A}_1)=5\text{ (Area }{A}_2)$, then $k$ is equal to :

Unlocking the Area Under the Curve

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric puzzle.

We have two regions, $A_1$ and $A_2$, and a mysterious constant $k$ that binds them together. Let us embark on this journey to find $k$ by understanding the soul of these equations.

Decoding the Parabolic Dance

First, let us look at $A_1=\{(x,y):|x|\le y^2,|x|+2y\le 8\}$. The absolute value $|x|\le y^2$ is a classic. It tells us that for any given height $y$, $x$ can range from $-y^2$ to $y^2$.

This is a parabola opening along the positive $y$-axis, but because of the $|x|$, it is perfectly mirrored across the $y$-axis. The second condition, $|x|+2y\le 8$, introduces a linear constraint.

If we look at the right side where $x\ge 0$, this becomes $x+2y\le 8$, or $x\le 8-2y$. This is a line with a negative slope.

The region $A_1$ is the intersection of these two constraints. Because of the symmetry, we can focus entirely on the right half ($x\ge 0$) and simply double our result at the end. This is the first secret to efficiency in JEE problems: never do double the work if you can use symmetry.

The Intersection Point

To find the area, we must know where the parabola and the line meet. We set $x=y^2$ and $x=8-2y$.

Equating them gives us the quadratic equation $y^2=8-2y$, which rearranges to:

Factoring this, we get $(y+4)(y-2)=0$. Since we are working in the upper half-plane ($y\ge 0$), we discard the negative root and find $y=2$.

At $y=2$, $x=2^2=4$. So, the intersection point is $(4,2)$. This point is the pivot of our integration.

The Art of Integration

Now, we calculate the area of the right half. We could integrate with respect to $x$, but that would require splitting the integral into two parts because the upper boundary changes.

Instead, let us integrate with respect to $y$. This is the elegant path. From $y=0$ to $y=2$, our horizontal strip is bounded by the parabola $x=y^2$.

From $y=2$ to $y=4$ (where the line hits the $y$-axis), the strip is bounded by the line $x=8-2y$. The area of the right half is given by the sum of these two integrals:

The first integral is:

The second integral is:

Adding these, we get $\frac{8}{3}+4=\frac{20}{3}$. Doubling this for the full region, we find $Area(A_1)=\frac{40}{3}$.

The Diamond of $A_2$ and the Final Victory

Finally, we turn to $A_2=\{(x,y):|x|+|y|\le k\}$. This is a beautiful shape. In each quadrant, it forms a line segment, creating a square rotated by $45^{\circ }$ with vertices at $(\pm k,0)$ and $(0,\pm k)$.

The diagonals are of length $2k$. The area of this square is:

We are given the relation $27\times Area(A_1)=5\times Area(A_2)$. Substituting our values, we get:

This simplifies to $9\times 40=10k^2$, or $360=10k^2$. Thus, $k^2=36$, and since $k$ must be positive, $k=6$.

We have arrived at the destination. The beauty of this problem lies not just in the final number, but in the symmetry and the geometric intuition that guided us there. Keep practicing, keep visualizing, and the math will always reveal its secrets.