Find the area bounded by the parabola `y ^(2) =4ax,` latus-rectum and x-axis.

Answer

To find the area bounded by the parabola \( y^2 = 4ax \), the latus rectum, and the x-axis, we can follow these steps:

### Step 1: Understand the Parabola and Latus Rectum
The given parabola is \( y^2 = 4ax \). The vertex of this parabola is at the origin (0,0), and the focus is at the point \( (a, 0) \). The latus rectum is a line segment that passes through the focus and is perpendicular to the axis of symmetry of the parabola. The endpoints of the latus rectum can be found by substituting \( x = a \) into the equation of the parabola.

### Step 2: Find the Points of the Latus Rectum
Substituting \( x = a \) into the parabola's equation:
\[ y^2 = 4a(a) = 4a^2 \]
Thus, \( y = \pm 2a \). Therefore, the endpoints of the latus rectum are \( (a, 2a) \) and \( (a, -2a) \).

### Step 3: Set Up the Integral for the Area
The area bounded by the parabola, the latus rectum, and the x-axis can be calculated using integration. We will integrate the upper half of the parabola from \( x = 0 \) to \( x = a \):
\[ \text{Area} = \int_{0}^{a} \left( \sqrt{4ax} - 0 \right) \, dx \]
Here, \( \sqrt{4ax} \) represents the upper half of the parabola, and \( 0 \) represents the x-axis.

### Step 4: Simplify the Integral
The integral becomes:
\[ \text{Area} = \int_{0}^{a} 2\sqrt{ax} \, dx \]
We can factor out the constant \( 2\sqrt{a} \):
\[ \text{Area} = 2\sqrt{a} \int_{0}^{a} \sqrt{x} \, dx \]

### Step 5: Evaluate the Integral
The integral \( \int \sqrt{x} \, dx \) can be evaluated as follows:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Thus,
\[ \text{Area} = 2\sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{a} = 2\sqrt{a} \cdot \frac{2}{3} \cdot a^{3/2} \]
Calculating this gives:
\[ \text{Area} = \frac{4}{3} a^2 \]

### Final Answer
The area bounded by the parabola \( y^2 = 4ax \), the latus rectum, and the x-axis is:
\[ \text{Area} = \frac{4}{3} a^2 \]