Let S and S' be the foci of the ellipse and B be any one of the extremities of its minor axis. If `DeltaS'BS=8sq.` units, then the length of a latus rectum of the ellipse is

Options

A. `2sqrt(2)`
B. `2`
C. `4`
D. `4sqrt(2)`

Answer

To find the length of the latus rectum of the ellipse given that the area of triangle \( S'BS \) is 8 square units, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the foci of the ellipse be \( S(Ae, 0) \) and \( S'(-Ae, 0) \), where \( A \) is the semi-major axis and \( e \) is the eccentricity.
- The extremity of the minor axis \( B \) will be at \( (0, B) \), where \( B \) is the semi-minor axis.

2. **Area of Triangle**:
- The area of triangle \( S'BS \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- The base of the triangle is the distance between the foci \( S \) and \( S' \), which is \( 2Ae \).
- The height of the triangle is the distance from point \( B \) to the x-axis, which is \( B \).

3. **Setting Up the Equation**:
- Given that the area of triangle \( S'BS = 8 \) square units, we can set up the equation:
\[ \frac{1}{2} \times (2Ae) \times B = 8 \]
- Simplifying this, we have:
\[ AeB = 8 \]

4. **Using the Eccentricity**:
- The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} \]
- Substituting \( e \) into the area equation gives:
\[ AB \sqrt{1 - \frac{B^2}{A^2}} = 8 \]

5. **Squaring Both Sides**:
- Squaring both sides to eliminate the square root:
\[ A^2B^2 \left(1 - \frac{B^2}{A^2}\right) = 64 \]
- Expanding this gives:
\[ A^2B^2 - B^4 = 64 \]

6. **Rearranging the Equation**:
- Rearranging gives us a quadratic in \( B^2 \):
\[ B^4 - A^2B^2 + 64 = 0 \]

7. **Finding the Roots**:
- Using the quadratic formula \( B^2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -A^2, c = 64 \):
\[ B^2 = \frac{A^2 \pm \sqrt{A^4 - 256}}{2} \]

8. **Condition for Real Roots**:
- For \( B^2 \) to be real and positive, we need \( A^4 - 256 \geq 0 \):
\[ A^4 \geq 256 \implies A^2 \geq 16 \implies A \geq 4 \]

9. **Calculating Latus Rectum**:
- The length of the latus rectum \( L \) of the ellipse is given by:
\[ L = \frac{2B^2}{A} \]
- Substituting \( B^2 = 8 \) and \( A = 4 \):
\[ L = \frac{2 \times 8}{4} = 4 \]

### Final Answer:
The length of the latus rectum of the ellipse is \( 4 \) units.