If the area bounded by `y^(2)=4ax and x^(2)=4ay` is `(64)/(3)` square units, then the positive value of a is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem of finding the positive value of \( a \) such that the area bounded by the curves \( y^2 = 4ax \) and \( x^2 = 4ay \) is \( \frac{64}{3} \) square units, we can follow these steps:

### Step 1: Identify the curves
The equations \( y^2 = 4ax \) and \( x^2 = 4ay \) represent two parabolas. The first parabola opens to the right and the second one opens upwards.

### Step 2: Find points of intersection
To find the points of intersection, we can substitute \( x \) from one equation into the other.

From \( y^2 = 4ax \), we can express \( x \) as:
\[ x = \frac{y^2}{4a} \]
Substituting this into the second equation \( x^2 = 4ay \):
\[ \left(\frac{y^2}{4a}\right)^2 = 4ay \]
This simplifies to:
\[ \frac{y^4}{16a^2} = 4ay \]
Multiplying through by \( 16a^2 \):
\[ y^4 = 64a^3y \]
Dividing both sides by \( y \) (assuming \( y \neq 0 \)):
\[ y^3 = 64a^3 \]
Thus, we find:
\[ y = 4a \]
Substituting \( y = 4a \) back into \( x = \frac{y^2}{4a} \):
\[ x = \frac{(4a)^2}{4a} = 4a \]
So the points of intersection are \( (4a, 4a) \) and the origin \( (0, 0) \).

### Step 3: Set up the area integral
The area \( A \) between the curves from \( 0 \) to \( 4a \) can be calculated as:
\[ A = \int_{0}^{4a} \left( \sqrt{4ax} - \frac{x^2}{4a} \right) dx \]
where \( \sqrt{4ax} \) is the upper curve and \( \frac{x^2}{4a} \) is the lower curve.

### Step 4: Compute the integral
Calculating the integral:
\[ A = \int_{0}^{4a} \left( 2\sqrt{ax} - \frac{x^2}{4a} \right) dx \]

1. **Integral of \( 2\sqrt{ax} \)**:
\[ = 2\int_{0}^{4a} \sqrt{ax} \, dx = 2\cdot \frac{2}{3} \cdot (4a)^{3/2} = \frac{8}{3} \cdot (4a)^{3/2} \]

2. **Integral of \( \frac{x^2}{4a} \)**:
\[ = \frac{1}{4a} \cdot \frac{x^3}{3} \bigg|_{0}^{4a} = \frac{1}{4a} \cdot \frac{(4a)^3}{3} = \frac{64a^2}{12} = \frac{16a^2}{3} \]

Putting it all together:
\[ A = \frac{8}{3} \cdot (4a)^{3/2} - \frac{16a^2}{3} \]
This simplifies to:
\[ A = \frac{8}{3} \cdot 8a^{3/2} - \frac{16a^2}{3} = \frac{64a^{3/2}}{3} - \frac{16a^2}{3} \]

### Step 5: Set the area equal to \( \frac{64}{3} \)
Setting the area equal to \( \frac{64}{3} \):
\[ \frac{64a^{3/2}}{3} - \frac{16a^2}{3} = \frac{64}{3} \]
Multiplying through by \( 3 \) to eliminate the denominator:
\[ 64a^{3/2} - 16a^2 = 64 \]
Rearranging gives:
\[ 64a^{3/2} - 16a^2 - 64 = 0 \]

### Step 6: Solve for \( a \)
Dividing the entire equation by \( 16 \):
\[ 4a^{3/2} - a^2 - 4 = 0 \]
Let \( x = a^{1/2} \) (thus \( a = x^2 \)):
\[ 4x^3 - x^4 - 4 = 0 \]
Rearranging gives:
\[ x^4 - 4x^3 + 4 = 0 \]
Factoring or using the Rational Root Theorem, we find \( x = 2 \) is a root. Thus:
\[ (x - 2)(x^3 - 2x^2 - 2x - 2) = 0 \]
The cubic can be solved, but we are primarily interested in \( x = 2 \).

### Step 7: Find \( a \)
Since \( x = a^{1/2} \), we have:
\[ a^{1/2} = 2 \implies a = 4 \]

### Conclusion
The positive value of \( a \) is:
\[ \boxed{4} \]