Area bounded by the curve `y^(2)=16x` and line y = mx is `(2)/(3)`, then m is equal to

Options

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

Explanation

To find the value of \( m \) such that the area bounded by the curve \( y^2 = 16x \) and the line \( y = mx \) is \( \frac{2}{3} \), we can follow these steps:

### Step 1: Identify the curves
The curve \( y^2 = 16x \) is a parabola that opens to the right with its vertex at the origin (0,0). The line \( y = mx \) is a straight line passing through the origin.

### Step 2: Find the points of intersection
To find the points of intersection between the parabola and the line, we substitute \( y = mx \) into the equation of the parabola:

\[ (mx)^2 = 16x \]

This simplifies to:

\[ m^2x^2 - 16x = 0 \]

Factoring out \( x \):

\[ x(m^2x - 16) = 0 \]

This gives us two solutions:

1. \( x = 0 \)
2. \( x = \frac{16}{m^2} \)

### Step 3: Set up the area integral
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{16}{m^2} \) can be expressed as:

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

### Step 4: Simplify the integral
First, we simplify the integrand:

\[ \sqrt{16x} = 4\sqrt{x} \]

Thus, the area becomes:

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

### Step 5: Calculate the integral
Now we can compute the integral:

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

Calculating each part:

1. For \( \int 4\sqrt{x} \, dx \):

\[ \int 4\sqrt{x} \, dx = \frac{4 \cdot 2}{3} x^{3/2} = \frac{8}{3} x^{3/2} \]

Evaluating from 0 to \( \frac{16}{m^2} \):

\[ \left[ \frac{8}{3} \left( \frac{16}{m^2} \right)^{3/2} \right] = \frac{8}{3} \cdot \frac{64}{m^3} = \frac{512}{3m^3} \]

2. For \( \int mx \, dx \):

\[ \int mx \, dx = \frac{m}{2} x^2 \]

Evaluating from 0 to \( \frac{16}{m^2} \):

\[ \left[ \frac{m}{2} \left( \frac{16}{m^2} \right)^2 \right] = \frac{m}{2} \cdot \frac{256}{m^4} = \frac{128}{m^3} \]

### Step 6: Combine the results
Now we can combine the results to find the total area:

\[ A = \frac{512}{3m^3} - \frac{128}{m^3} = \frac{512 - 384}{3m^3} = \frac{128}{3m^3} \]

### Step 7: Set the area equal to \( \frac{2}{3} \)
We set the area equal to \( \frac{2}{3} \):

\[ \frac{128}{3m^3} = \frac{2}{3} \]

### Step 8: Solve for \( m \)
Cross-multiplying gives:

\[ 128 = 2m^3 \]

Dividing both sides by 2:

\[ m^3 = 64 \]

Taking the cube root:

\[ m = 4 \]

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

\[ \boxed{4} \]