The area of the region bounded by the parabola `y^(2)=4ax` and the line y=mx is

Options

A. `(8a^(2))/(3m^(3))`
B. `(8m^(2))/(3a^(3))`
C. `(8a^(2))/3`
D. `(8a^(2)m^(3))/3`

Answer

To find the area of the region bounded by the parabola \( y^2 = 4ax \) and the line \( y = mx \), we will follow these steps:

### Step 1: Find the points of intersection
To find the area, we first need to determine the points where the parabola and the line intersect. We can do this by substituting \( y = mx \) into the equation of the parabola.

1. Substitute \( y = mx \) into \( y^2 = 4ax \):
\[ (mx)^2 = 4ax \]
This simplifies to:
\[ m^2x^2 = 4ax \]
Rearranging gives:
\[ m^2x^2 - 4ax = 0 \]
Factoring out \( x \):
\[ x(m^2x - 4a) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad m^2x = 4a \quad \Rightarrow \quad x = \frac{4a}{m^2} \]

### Step 2: Find the corresponding y-values
Now we find the y-coordinates of the intersection points by substituting back into the line equation \( y = mx \).

1. For \( x = 0 \):
\[ y = m(0) = 0 \quad \Rightarrow \quad (0, 0) \]

2. For \( x = \frac{4a}{m^2} \):
\[ y = m\left(\frac{4a}{m^2}\right) = \frac{4a}{m} \]
Thus, the second intersection point is:
\[ \left(\frac{4a}{m^2}, \frac{4a}{m}\right) \]

### Step 3: Set up the integral for the area
The area \( A \) between the curve and the line from \( x = 0 \) to \( x = \frac{4a}{m^2} \) can be expressed as:
\[ A = \int_{0}^{\frac{4a}{m^2}} \left( \text{upper function} - \text{lower function} \right) \, dx \]
In this case, the upper function is the parabola \( y = \sqrt{4ax} \) and the lower function is the line \( y = mx \).

### Step 4: Calculate the integral
Thus, we have:
\[ A = \int_{0}^{\frac{4a}{m^2}} \left( \sqrt{4ax} - mx \right) \, dx \]

Now, we can compute the integral:
1. The integral of \( \sqrt{4ax} \):
\[ \int \sqrt{4ax} \, dx = \int 2\sqrt{a} \sqrt{x} \, dx = \frac{4}{3} \cdot 2\sqrt{a} x^{3/2} = \frac{8\sqrt{a}}{3} x^{3/2} \]

2. The integral of \( mx \):
\[ \int mx \, dx = \frac{m}{2} x^2 \]

Putting it all together:
\[ A = \left[ \frac{8\sqrt{a}}{3} x^{3/2} - \frac{m}{2} x^2 \right]_{0}^{\frac{4a}{m^2}} \]

### Step 5: Evaluate the definite integral
Substituting the limits:
1. For \( x = \frac{4a}{m^2} \):
\[ A = \frac{8\sqrt{a}}{3} \left(\frac{4a}{m^2}\right)^{3/2} - \frac{m}{2} \left(\frac{4a}{m^2}\right)^2 \]
Simplifying gives:
\[ = \frac{8\sqrt{a}}{3} \cdot \frac{8a^{3/2}}{m^3} - \frac{m}{2} \cdot \frac{16a^2}{m^4} \]
\[ = \frac{64a^2}{3m^3} - \frac{8a^2}{m^3} = \frac{64a^2 - 24a^2}{3m^3} = \frac{40a^2}{3m^3} \]

Thus, the area of the region bounded by the parabola and the line is:
\[ A = \frac{40a^2}{3m^3} \]