If the area bounded by the parabola `y^(2)=16` ax and the line y = 4 mx is `(a^(2))/(12)` sq unit then using integration find the value of m.

Answer

To solve the problem, we need to find the value of \( m \) given that the area bounded by the parabola \( y^2 = 16Ax \) and the line \( y = 4mx \) is \( \frac{A^2}{12} \) square units. We will use integration to find this area.

### Step 1: Find the points of intersection
First, we need to find the points where the parabola and the line intersect. We substitute \( y = 4mx \) into the parabola's equation:

\[ (4mx)^2 = 16Ax \]

This simplifies to:

\[ 16m^2x^2 = 16Ax \]

Dividing both sides by 16 (assuming \( x \neq 0 \)) gives:

\[ m^2x^2 = Ax \]

Rearranging this, we have:

\[ m^2x^2 - Ax = 0 \]

Factoring out \( x \):

\[ x(m^2x - A) = 0 \]

This gives us two solutions:

1. \( x = 0 \)
2. \( m^2x - A = 0 \) which leads to \( x = \frac{A}{m^2} \)

Thus, the points of intersection are \( (0, 0) \) and \( \left(\frac{A}{m^2}, 4m\frac{A}{m^2}\right) = \left(\frac{A}{m^2}, \frac{4A}{m}\right) \).

### Step 2: Set up the integral for the area
The area \( A \) between the parabola and the line from \( x = 0 \) to \( x = \frac{A}{m^2} \) is given by:

\[ \text{Area} = \int_0^{\frac{A}{m^2}} \left( \sqrt{16Ax} - 4mx \right) \, dx \]

### Step 3: Evaluate the integral
We can split the integral into two parts:

\[ \text{Area} = \int_0^{\frac{A}{m^2}} \sqrt{16Ax} \, dx - \int_0^{\frac{A}{m^2}} 4mx \, dx \]

#### Evaluating the first integral:
Let \( I_1 = \int_0^{\frac{A}{m^2}} \sqrt{16Ax} \, dx \).

Using the substitution \( u = 16Ax \), we have \( du = 16A \, dx \) or \( dx = \frac{du}{16A} \).

The limits change as follows:
- When \( x = 0 \), \( u = 0 \)
- When \( x = \frac{A}{m^2} \), \( u = 16A \cdot \frac{A}{m^2} = \frac{16A^2}{m^2} \)

Thus,

\[ I_1 = \int_0^{\frac{16A^2}{m^2}} \sqrt{u} \cdot \frac{du}{16A} = \frac{1}{16A} \cdot \frac{2}{3} u^{3/2} \bigg|_0^{\frac{16A^2}{m^2}} = \frac{1}{16A} \cdot \frac{2}{3} \left( \frac{16A^2}{m^2} \right)^{3/2} \]

Calculating this gives:

\[ = \frac{1}{16A} \cdot \frac{2}{3} \cdot \frac{64A^3}{m^3} = \frac{8A^2}{3m^3} \]

#### Evaluating the second integral:
Let \( I_2 = \int_0^{\frac{A}{m^2}} 4mx \, dx \).

This evaluates to:

\[ I_2 = 4m \cdot \frac{x^2}{2} \bigg|_0^{\frac{A}{m^2}} = 2m \cdot \left( \frac{A}{m^2} \right)^2 = \frac{2A^2}{m^3} \]

### Step 4: Combine the results
Now, the total area is:

\[ \text{Area} = I_1 - I_2 = \frac{8A^2}{3m^3} - \frac{2A^2}{m^3} = \frac{8A^2 - 6A^2}{3m^3} = \frac{2A^2}{3m^3} \]

### Step 5: Set the area equal to \( \frac{A^2}{12} \)
According to the problem, this area is equal to \( \frac{A^2}{12} \):

\[ \frac{2A^2}{3m^3} = \frac{A^2}{12} \]

Dividing both sides by \( A^2 \) (assuming \( A \neq 0 \)) gives:

\[ \frac{2}{3m^3} = \frac{1}{12} \]

Cross-multiplying yields:

\[ 2 \cdot 12 = 3m^3 \implies 24 = 3m^3 \implies m^3 = 8 \implies m = 2 \]

### Final Answer
Thus, the value of \( m \) is:

\[ \boxed{2} \]