Find the area bounded by the parabola `y^2 = 4ax` and the line `y = 2ax`.

Options

A. 3a sq. units
B. `(1)/(3a)` sq. units
C. `(2)/(3a)` sq. units
D. None of these

Answer

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

### Step 1: Find the Points of Intersection
To find the area bounded by the two curves, we first need to determine where they intersect. We can do this by substituting \( y = 2ax \) into the equation of the parabola.

Given:
1. \( y^2 = 4ax \)
2. \( y = 2ax \)

Substituting \( y \) into the parabola's equation:
\[ (2ax)^2 = 4ax \]
This simplifies to:
\[ 4a^2x^2 = 4ax \]
Dividing both sides by \( 4a \) (assuming \( a \neq 0 \)):
\[ a x^2 = x \]
Rearranging gives:
\[ ax^2 - x = 0 \]
Factoring out \( x \):
\[ x(ax - 1) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x = \frac{1}{a} \]

### Step 2: Find Corresponding y-values
Now, we find the corresponding \( y \)-values for these \( x \)-values using \( y = 2ax \):
- For \( x = 0 \):
\[ y = 2a(0) = 0 \quad \Rightarrow \quad (0, 0) \]
- For \( x = \frac{1}{a} \):
\[ y = 2a\left(\frac{1}{a}\right) = 2 \quad \Rightarrow \quad \left(\frac{1}{a}, 2\right) \]

### Step 3: Set Up the Integral for Area
The area \( A \) bounded by the curves can be found using the integral from \( x = 0 \) to \( x = \frac{1}{a} \) of the top curve minus the bottom curve. Here, the parabola is above the line.

The area is given by:
\[ A = \int_{0}^{\frac{1}{a}} (y_{\text{parabola}} - y_{\text{line}}) \, dx \]
Where:
- \( y_{\text{parabola}} = 2\sqrt{ax} \) (from \( y^2 = 4ax \))
- \( y_{\text{line}} = 2ax \)

Thus, the area becomes:
\[ A = \int_{0}^{\frac{1}{a}} \left(2\sqrt{ax} - 2ax\right) \, dx \]

### Step 4: Simplify the Integral
Factor out the constant:
\[ A = 2 \int_{0}^{\frac{1}{a}} \left(\sqrt{ax} - ax\right) \, dx \]
This can be rewritten as:
\[ A = 2 \int_{0}^{\frac{1}{a}} \left(\sqrt{a} x^{1/2} - ax\right) \, dx \]

### Step 5: Compute the Integral
Now we can compute the integral:
1. The integral of \( \sqrt{a} x^{1/2} \):
\[ \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \]
2. The integral of \( ax \):
\[ \int ax \, dx = \frac{a}{2} x^2 \]

Now substituting back:
\[ A = 2 \left[ \sqrt{a} \cdot \frac{2}{3} x^{3/2} - a \cdot \frac{1}{2} x^2 \right]_{0}^{\frac{1}{a}} \]
Calculating at the limits:
\[ = 2 \left[ \sqrt{a} \cdot \frac{2}{3} \left(\frac{1}{a}\right)^{3/2} - a \cdot \frac{1}{2} \left(\frac{1}{a}\right)^2 \right] \]
\[ = 2 \left[ \sqrt{a} \cdot \frac{2}{3} \cdot \frac{1}{a^{3/2}} - a \cdot \frac{1}{2} \cdot \frac{1}{a^2} \right] \]
\[ = 2 \left[ \frac{2}{3\sqrt{a}} - \frac{1}{2a} \right] \]

### Step 6: Combine Terms
Finding a common denominator:
\[ = 2 \left[ \frac{4 - 3}{6a} \right] = \frac{2}{6a} = \frac{1}{3a} \]

### Final Answer
Thus, the area bounded by the parabola and the line is:
\[ \boxed{\frac{1}{3a}} \text{ square units} \]