Animated Solution for Mathematics - Definite Integration: If the area enclosed between the curves $y=k{x}^2$ and $x=k{y}^2,(k>0)$, is 1 square unit. Then k is:

Imagine standing in the first quadrant of the Cartesian plane. You are looking at two elegant curves: $y=k{x}^2$ and $x=k{y}^2$.

Because $k>0$, the first curve is a classic parabola opening upwards, hugging the $y$-axis. The second curve is its mirror image, a parabola opening rightwards, hugging the $x$-axis.

Together, they create a beautiful, leaf-shaped region trapped between them. We are tasked with finding the value of $k$ such that the area of this leaf is exactly $1$ square unit.

Before we calculate the area, we must define the boundaries. We solve the system of equations: $y=k{x}^2$ and $x=k{y}^2$.

Substituting the first into the second, we get $x=k(k{x}^2{)}^2$, which simplifies to $x=k^3{x}^4$. This gives us the equation:

The solutions are $x=0$ and $x=\frac{1}{k}$. Thus, our region of interest is bounded by $x=0$ and $x=\frac{1}{k}$.

Standard Forms

While you could certainly set up a definite integral to find this area, the JEE Advanced exam rewards those who know the shortcuts. Let us transform our equations into the standard forms of parabolas: $y^2=4ax$ and $x^2=4by$.

Our given equations are $y=k{x}^2$ and $x=k{y}^2$. Rearranging these, we get:

Now, compare these to the standard forms. For $x^2=\frac{1}{k}y$, we see that $4b=\frac{1}{k}$, which means $b=\frac{1}{4k}$. Similarly, for $y^2=\frac{1}{k}x$, we see that $4a=\frac{1}{k}$, which means $a=\frac{1}{4k}$.

There is a beautiful, well-known formula for the area enclosed between these two parabolas:

Let us substitute our values for $a$ and $b$ into this formula:

Simplifying this expression, we obtain:

The problem states that this area is $1$. So, we set $\frac{1}{3k^2}=1$, which leads to $3k^2=1$, or $k^2=\frac{1}{3}$.

Taking the square root, we find $k=\pm \frac{1}{\sqrt{3}}$. Since the problem explicitly states $k>0$, we discard the negative root.

Our final answer is $k=\frac{1}{\sqrt{3}}$.