Find the area of the parabola `y^2=4a x`bounded by its latus rectum.

Answer

To find the area of the parabola given by the equation \( y^2 = 4ax \) bounded by its latus rectum, we can follow these steps:

### Step 1: Understand the Parabola and Latus Rectum
The equation \( y^2 = 4ax \) represents a rightward-opening parabola. The latus rectum of this parabola is a vertical line segment that passes through the focus of the parabola. For the parabola \( y^2 = 4ax \), the focus is at the point \( (a, 0) \).

### Step 2: Identify the Boundaries
The latus rectum is defined by the vertical lines \( x = a \) and \( x = 0 \) (the y-axis). We will find the area between these two lines.

### Step 3: Express \( y \) in Terms of \( x \)
From the equation \( y^2 = 4ax \), we can express \( y \) as:
\[ y = \sqrt{4ax} = 2\sqrt{ax} \]
This gives us the upper half of the parabola. The lower half would be \( y = -2\sqrt{ax} \).

### Step 4: Set Up the Integral
To find the area bounded by the parabola and the latus rectum, we will integrate \( y \) from \( x = 0 \) to \( x = a \):
\[ \text{Area} = \int_{0}^{a} y \, dx = \int_{0}^{a} 2\sqrt{ax} \, dx \]

### Step 5: Solve the Integral
Now we will compute the integral:
\[ \text{Area} = 2\int_{0}^{a} \sqrt{ax} \, dx \]
We can factor out the constant \( \sqrt{a} \):
\[ = 2\sqrt{a} \int_{0}^{a} \sqrt{x} \, dx \]
The integral \( \int \sqrt{x} \, dx \) can be computed as follows:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Now we evaluate this from \( 0 \) to \( a \):
\[ = 2\sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{a} = 2\sqrt{a} \left( \frac{2}{3} a^{3/2} - 0 \right) \]
\[ = 2\sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{4}{3} a^2 \]

### Step 6: Final Result
Thus, the area of the parabola \( y^2 = 4ax \) bounded by its latus rectum is:
\[ \text{Area} = \frac{4}{3} a^2 \text{ square units} \]