The area enclosed between the curves y = ax² and x = ay² (a 0) is 1 sq. unit, then the value of a

Q: Why can't I just integrate the difference between the two curves?

You absolutely can! Integrating $\int_{0}^{1/a} (\sqrt{x/a} - ax^2) dx$ will yield the same result. However, in competitive exams like JEE, time is your most valuable asset. Using the shortcut formula $\frac{16AB}{3}$ allows you to bypass the integration steps entirely, reducing the chance of calculation errors.

Q: How do I know which parabola is which?

Think about the variables. $y = ax^2$ means for every $x$, there is a unique $y$, which is the definition of a vertical parabola. Conversely, $x = ay^2$ means for every $y$, there is a unique $x$, defining a horizontal parabola. Always sketch them to confirm the orientation.

Q: Why do we reject the negative root when solving $a^2 = 1/3$?

Great catch! The problem explicitly states that $a > 0$. When we solve $a^2 = 1/3$, we get $a = \pm 1/\sqrt{3}$. Since the condition $a > 0$ is a constraint given in the problem, we must discard the negative solution to satisfy the physical reality of the curves defined.

Question

The area enclosed between the curves $y = ax^2$ and $x = ay^2$ ($a > 0$) is 1 sq. unit, then the value of $a$ is

Options:

(A) $1/\sqrt{3}$

(B) $1/2$

(C) 1

(D) $1/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, JEE warriors! Today, we are going to explore the elegant dance of two parabolas. Imagine you are standing on the Cartesian plane, looking at two curves: $y = ax^2$ and $x = ay^2$, where $a > 0$.

These are not just lines on a graph; they are the heartbeat of coordinate geometry. The first, $y = ax^2$, is a classic upward-opening parabola, while the second, $x = ay^2$, is a rightward-opening parabola.

Together, they intersect at the origin $(0,0)$ and at the point $(\frac{1}{a}, \frac{1}{a})$, trapping a beautiful, leaf-like region between them. Our mission is to find the value of $a$ such that the area of this region is exactly $1$ square unit.

### The Power of the Shortcut

While you could certainly set up a definite integral to calculate this area, we want to be smarter, not just harder. In the JEE, time is your most precious resource.

There is a powerful, time-tested shortcut for this exact scenario. The area enclosed between two standard parabolas, $y^2 = 4Ax$ and $x^2 = 4By$, is given by the formula:

$$	ext{Area} = \frac{16AB}{3}$$

This formula is a gift, derived from the scaling properties of parabolas, and it will save you precious minutes during the examination.

### The Algebraic Transformation

To wield this weapon, we must first mold our given equations into the standard forms. Let us take $x = ay^2$ and rewrite it as $y^2 = \frac{1}{a}x$.

Similarly, we rewrite $y = ax^2$ as $x^2 = \frac{1}{a}y$. Now, we compare these with our standard forms: $y^2 = 4Ax$ and $x^2 = 4By$.

By matching the coefficients, we see that $4A = \frac{1}{a}$, which means $A = \frac{1}{4a}$. By the same logic, $4B = \frac{1}{a}$, so $B = \frac{1}{4a}$.

### The Final Calculation

Now, we substitute these values into our area formula:

$$	ext{Area} = \frac{16}{3} \cdot A \cdot B = \frac{16}{3} \cdot \left( \frac{1}{4a} ight) \cdot \left( \frac{1}{4a} ight)$$

Simplifying this, we get:

$$	ext{Area} = \frac{16}{3} \cdot \frac{1}{16a^2} = \frac{1}{3a^2}$$

We are given that this area is $1$ square unit. So, we set $\frac{1}{3a^2} = 1$, which leads us to $3a^2 = 1$, or $a^2 = \frac{1}{3}$.

Taking the square root, we get $a = \pm \frac{1}{\sqrt{3}}$. But wait! Remember the constraint $a > 0$? We must reject the negative root.

Thus, our final, elegant answer is **$a = \frac{1}{\sqrt{3}}$**.

Animated Solution for Mathematics - Definite Integration: The area enclosed between the curves $y = a x^{2}$ and $x = a y^{2}$ ( $a > 0$ ) is 1 sq. unit, then the value of $a$ is

Select Answer:

Visualized Solution

Visualizing the Curves

The Enclosed Area

The Shortcut Formula

Converting to Standard Form

Identifying Coefficients $A$ and $B$

Substituting into Area Formula

Simplifying the Expression

Equating to Given Area

Final Value of $a$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Power of the Shortcut

The Algebraic Transformation

The Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area enclosed between the curves y=ax2 and x=ay2 (a>0) is 1 sq. unit, then the value of a is