If the length of the semi-major axis of an ellipse is 68 and e = 1/2, then the area of the rectangle formed by joining the vertices of latusrecta of the ellipse is equal to

Options

A. 69930
B. 6935
C. 6936
D. 3696

Answer

To solve the problem step by step, we need to find the area of the rectangle formed by joining the vertices of the latus rectum of the ellipse given the semi-major axis and eccentricity.

### Step 1: Identify the given values
- Length of the semi-major axis (B) = 68
- Eccentricity (e) = 1/2

### Step 2: Use the relationship between semi-major axis, semi-minor axis, and eccentricity
The relationship is given by:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} \]
Where:
- A = semi-major axis
- B = semi-minor axis
- e = eccentricity

### Step 3: Substitute the known values into the equation
Substituting the values of e and B into the equation:
\[ \frac{1}{2} = \sqrt{1 - \frac{68^2}{A^2}} \]

### Step 4: Square both sides to eliminate the square root
\[ \left(\frac{1}{2}\right)^2 = 1 - \frac{68^2}{A^2} \]
\[ \frac{1}{4} = 1 - \frac{4624}{A^2} \]

### Step 5: Rearrange the equation to solve for A^2
\[ \frac{4624}{A^2} = 1 - \frac{1}{4} \]
\[ \frac{4624}{A^2} = \frac{3}{4} \]

### Step 6: Cross-multiply to find A^2
\[ 4624 \cdot 4 = 3A^2 \]
\[ 18496 = 3A^2 \]
\[ A^2 = \frac{18496}{3} \]

### Step 7: Calculate A
\[ A = \sqrt{\frac{18496}{3}} \]

### Step 8: Find the area of the rectangle formed by the vertices of the latus rectum
The area of the rectangle formed by the vertices of the latus rectum is given by:
\[ \text{Area} = 2AE \times 2B \]

### Step 9: Calculate E using the eccentricity
Since e = 1/2, we have:
\[ E = e = \frac{1}{2} \]

### Step 10: Substitute A, E, and B into the area formula
Substituting the values:
\[ \text{Area} = 2 \times A \times \frac{1}{2} \times 2 \times 68 \]
\[ \text{Area} = 2A \times 68 \]

### Step 11: Substitute A from Step 7
Substituting A:
\[ \text{Area} = 2 \times \sqrt{\frac{18496}{3}} \times 68 \]

### Step 12: Calculate the final area
Calculating the area:
\[ \text{Area} = 2 \times 68 \times \sqrt{\frac{18496}{3}} \]
\[ \text{Area} = 68^2 \times \frac{4}{3} \]
\[ \text{Area} = 4624 \times \frac{4}{3} \]
\[ \text{Area} = 18496/3 \]
\[ \text{Area} = 6165.33 \text{ square units} \]

### Final Answer
The area of the rectangle formed by joining the vertices of the latus rectum of the ellipse is approximately **6165.33 square units**.

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