The area enclosed between the curves `y=ax^(2)` and `x=ay^(2)(a gt 0)` is 1 sq.unit. Then a =

Options

A. `(1)/(sqrt(3))`
B. `(2)/(sqrt(3))`
C. `(4)/(sqrt(3))`
D. `sqrt(3)`

Answer

To find the value of \( a \) such that the area enclosed between the curves \( y = ax^2 \) and \( x = ay^2 \) is 1 square unit, we can follow these steps:

### Step 1: Identify the curves
The equations of the curves are:
1. \( y = ax^2 \) (a parabola opening upwards)
2. \( x = ay^2 \) (a parabola opening to the right)

### Step 2: Find the points of intersection
To find the points of intersection, we can substitute \( y = ax^2 \) into \( x = ay^2 \):
\[ x = a(ax^2)^2 = a^3x^4 \]
Rearranging gives:
\[ a^3x^4 - x = 0 \]
Factoring out \( x \):
\[ x(a^3x^3 - 1) = 0 \]
This gives us \( x = 0 \) or \( a^3x^3 = 1 \). Solving for \( x \):
\[ x = \frac{1}{a} \]
Thus, the points of intersection are \( (0, 0) \) and \( \left(\frac{1}{a}, \frac{1}{a}\right) \).

### Step 3: Set up the area integral
The area \( A \) enclosed between the curves from \( x = 0 \) to \( x = \frac{1}{a} \) can be calculated as:
\[ A = \int_0^{\frac{1}{a}} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]
Where \( y_{\text{upper}} = ax^2 \) and \( y_{\text{lower}} = \sqrt{\frac{x}{a}} \).

### Step 4: Calculate the area
The area can be expressed as:
\[ A = \int_0^{\frac{1}{a}} \left(ax^2 - \sqrt{\frac{x}{a}}\right) \, dx \]

### Step 5: Evaluate the integral
Calculating the integral:
1. The integral of \( ax^2 \):
\[ \int ax^2 \, dx = \frac{a}{3}x^3 \bigg|_0^{\frac{1}{a}} = \frac{a}{3} \left(\frac{1}{a}\right)^3 = \frac{1}{3a^2} \]

2. The integral of \( \sqrt{\frac{x}{a}} \):
\[ \int \sqrt{\frac{x}{a}} \, dx = \sqrt{\frac{1}{a}} \cdot \frac{2}{3}x^{3/2} \bigg|_0^{\frac{1}{a}} = \sqrt{\frac{1}{a}} \cdot \frac{2}{3} \left(\frac{1}{a}\right)^{3/2} = \frac{2}{3a^{2}} \]

Putting it all together:
\[ A = \frac{1}{3a^2} - \frac{2}{3a^2} = -\frac{1}{3a^2} \]
Since we are looking for the area, we take the absolute value:
\[ A = \frac{1}{3a^2} \]

### Step 6: Set the area equal to 1 square unit
According to the problem, the area is given as 1 square unit:
\[ \frac{1}{3a^2} = 1 \]
Solving for \( a^2 \):
\[ a^2 = \frac{1}{3} \implies a = \frac{1}{\sqrt{3}} \]

### Conclusion
Thus, the value of \( a \) is:
\[ \boxed{\frac{1}{\sqrt{3}}} \]