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 solve the problem, we need to find the length of the latus rectum of the ellipse given that the area of triangle S'BS is 8 square units, where S and S' are the foci of the ellipse, and B is an extremity of its minor axis.

### Step-by-Step Solution:

1. **Understand the Geometry of the Ellipse**:
- The foci of the ellipse are located at S(-ae, 0) and S'(ae, 0).
- The extremities of the minor axis are at B(0, b) and B(0, -b).

2. **Area of Triangle S'BS**:
- The area of triangle S'BS can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Here, the base can be taken as the distance between the foci S and S', which is \(2ae\), and the height is the distance from B to the line joining S and S', which is \(b\).
- Therefore, the area can be expressed as:
\[ \text{Area} = \frac{1}{2} \times (2ae) \times b = aeb \]
- We know from the problem that this area is equal to 8 square units:
\[ aeb = 8 \]

3. **Relationship between a, b, and e**:
- The eccentricity \(e\) of the ellipse is defined as:
\[ e = \frac{c}{a} \]
- Where \(c = ae\) and \(b^2 = a^2(1 - e^2)\).

4. **Substituting for b**:
- From the geometry of the ellipse, we know that \(b = ae\).
- Substituting this into the area equation gives:
\[ ae(ae) = 8 \implies a^2e^2 = 8 \]

5. **Using the relationship between a and b**:
- From the properties of the ellipse, we also have:
\[ b^2 = a^2(1 - e^2) \]
- Substituting \(b = ae\) gives:
\[ (ae)^2 = a^2(1 - e^2) \implies a^2e^2 = a^2(1 - e^2) \]

6. **Solving for a and e**:
- From \(a^2e^2 = 8\) and \(a^2e^2 = a^2(1 - e^2)\), we can equate:
\[ 8 = a^2(1 - e^2) \implies 8 = a^2 - a^2e^2 \]
- Substituting \(a^2e^2 = 8\) gives:
\[ 8 = a^2 - 8 \implies a^2 = 16 \implies a = 4 \]

7. **Finding the Length of the Latus Rectum**:
- The length of the latus rectum \(L\) of the ellipse is given by:
\[ L = \frac{2b^2}{a} \]
- We already know \(b^2 = a^2e^2 = 8\), thus:
\[ L = \frac{2 \times 8}{4} = 4 \]

### Final Answer:
The length of the latus rectum of the ellipse is **4 units**.

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