Statement I- The area of region bounded parabola `y^(2)=4x and x^(2)=4y " is " 32/3` sq units.
Statement II- The area of region bounded by parabola `y^(2)=4ax and x^(2)=4by " is " 16/3 ab`.

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of Statement I.
B. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.
C. Statement I is true , Statement II is false
D. Statement I is false , Statement II is true

Answer

To solve the problem, we need to analyze the two statements regarding the areas bounded by the given parabolas.

### Step 1: Identify the equations of the parabolas
The first parabola is given by:
\[ y^2 = 4x \]
The second parabola is given by:
\[ x^2 = 4by \]

### Step 2: Find the points of intersection
To find the points of intersection, we can express \( y \) from the first equation and substitute it into the second equation.

From the first equation:
\[ y = \sqrt{4x} = 2\sqrt{x} \]

Substituting \( y \) into the second equation:
\[ x^2 = 4b(2\sqrt{x}) \]
\[ x^2 = 8b\sqrt{x} \]

Squaring both sides:
\[ x^4 = 64b^2x \]
Rearranging gives:
\[ x^4 - 64b^2x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 64b^2) = 0 \]

Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x^3 = 64b^2 \]
From \( x^3 = 64b^2 \), we find:
\[ x = (64b^2)^{1/3} = 4b^{2/3} \]

The points of intersection are:
1. \( (0, 0) \)
2. \( (4b^{2/3}, 2\sqrt{4b^{2/3}}) = (4b^{2/3}, 4b^{1/3}) \)

### Step 3: Set up the area integral
The area \( A \) between the curves can be found using the integral:
\[ A = \int_{0}^{4b^{2/3}} \left( \text{upper curve} - \text{lower curve} \right) \, dx \]

The upper curve is \( y = 2\sqrt{x} \) and the lower curve is \( y = \frac{x^2}{4b} \).

### Step 4: Calculate the area
The area can be expressed as:
\[ A = \int_{0}^{4b^{2/3}} \left( 2\sqrt{x} - \frac{x^2}{4b} \right) \, dx \]

Calculating the integral:
1. The integral of \( 2\sqrt{x} \):
\[ \int 2\sqrt{x} \, dx = \frac{4}{3} x^{3/2} \]
2. The integral of \( \frac{x^2}{4b} \):
\[ \int \frac{x^2}{4b} \, dx = \frac{x^3}{12b} \]

Thus, we have:
\[ A = \left[ \frac{4}{3} x^{3/2} - \frac{x^3}{12b} \right]_{0}^{4b^{2/3}} \]

Evaluating at the limits:
1. At \( x = 4b^{2/3} \):
\[ \frac{4}{3} (4b^{2/3})^{3/2} - \frac{(4b^{2/3})^3}{12b} \]
\[ = \frac{4}{3} (8b) - \frac{64b^2}{12b} \]
\[ = \frac{32b}{3} - \frac{16b}{3} = \frac{16b}{3} \]

2. At \( x = 0 \), both terms are zero.

Thus, the area is:
\[ A = \frac{16b}{3} \]

### Step 5: Conclusion
The area bounded by the two parabolas is:
\[ A = \frac{16}{3} ab \]
This confirms that **Statement II** is true.