Area bounded by the parabola `x^(2)=4y` and its latusrectum is

Options

A. `(4)/(3)`
B. `(8)/(3)`
C. `(16)/(3)`
D. `(5)/(3)`

Answer

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

### Step 1: Identify the parabola and its parameters
The given parabola is \( x^2 = 4y \). This is a standard form of a parabola that opens upwards. Here, we can identify that \( 4a = 4 \), which gives us \( a = 1 \).

### Step 2: Determine the latus rectum
The length of the latus rectum of a parabola \( x^2 = 4ay \) is given by \( 4a \). Since we have already found \( a = 1 \), the length of the latus rectum is:
\[ 4a = 4 \times 1 = 4 \]

### Step 3: Find the coordinates of the endpoints of the latus rectum
The endpoints of the latus rectum can be found at the points \( (2a, a) \) and \( (-2a, a) \). Substituting \( a = 1 \):
- Right endpoint: \( (2 \times 1, 1) = (2, 1) \)
- Left endpoint: \( (-2 \times 1, 1) = (-2, 1) \)

### Step 4: Set up the integral for the area
The area bounded by the parabola and the latus rectum can be calculated using definite integrals. The area can be expressed as:
\[ \text{Area} = \int_{-2}^{2} (y_{\text{parabola}} - y_{\text{line}}) \, dx \]
Here, \( y_{\text{parabola}} = \frac{x^2}{4} \) and \( y_{\text{line}} = 1 \) (the y-coordinate of the latus rectum).

### Step 5: Write the integral
The area can thus be written as:
\[ \text{Area} = \int_{-2}^{2} \left( \frac{x^2}{4} - 1 \right) \, dx \]

### Step 6: Calculate the integral
Calculating the integral:
\[ \text{Area} = \int_{-2}^{2} \left( \frac{x^2}{4} - 1 \right) \, dx = \int_{-2}^{2} \frac{x^2}{4} \, dx - \int_{-2}^{2} 1 \, dx \]

Calculating each part:
1. For \( \int_{-2}^{2} \frac{x^2}{4} \, dx \):
\[ = \frac{1}{4} \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{1}{4} \left( \frac{2^3}{3} - \frac{(-2)^3}{3} \right) = \frac{1}{4} \left( \frac{8}{3} + \frac{8}{3} \right) = \frac{1}{4} \cdot \frac{16}{3} = \frac{4}{3} \]

2. For \( \int_{-2}^{2} 1 \, dx \):
\[ = [x]_{-2}^{2} = 2 - (-2) = 4 \]

Combining these results:
\[ \text{Area} = \frac{4}{3} - 4 = \frac{4}{3} - \frac{12}{3} = -\frac{8}{3} \]
Since area cannot be negative, we take the absolute value:
\[ \text{Area} = \frac{8}{3} \]

### Final Answer
The area bounded by the parabola \( x^2 = 4y \) and its latus rectum is \( \frac{8}{3} \).

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