S and T are foci of an ellipse and B is an end of the minor axis , if STB is an equilateral triangle , the eccentricity of the ellipse , is

Options

A. `(1)/(4)`
B. `(1)/(3)`
C. `(1)/(2)`
D. `(2)/(3)`

Answer

To find the eccentricity of the ellipse given that S and T are the foci and B is an end of the minor axis, forming an equilateral triangle STB, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- Let the distance between the foci S and T be \(2c\).
- Let the length of the semi-major axis be \(a\).
- The length of the semi-minor axis is \(b\).
- Since B is an end of the minor axis, the coordinates of B can be taken as (0, b).

2. **Equilateral Triangle Properties**:
- Since triangle STB is equilateral, each angle is \(60^\circ\).
- The distance ST is equal to the distance SB and TB.

3. **Using Right Triangle Properties**:
- In triangle STB, we can drop a perpendicular from B to the line segment ST. Let's denote the foot of the perpendicular as O.
- The angle ∠STB = 60°, hence ∠SOB = 30° and ∠TOB = 30°.

4. **Using Trigonometric Ratios**:
- In triangle SOB, we can use the tangent function:
\[ \tan(60^\circ) = \frac{OB}{OS} \]
- Here, \(OB = b\) (the semi-minor axis) and \(OS = c\) (the distance from the center to a focus).

5. **Substituting Values**:
- We know that \(\tan(60^\circ) = \sqrt{3}\):
\[ \sqrt{3} = \frac{b}{c} \]
- Rearranging gives us:
\[ b = c\sqrt{3} \]

6. **Using the Relationship of a, b, and c**:
- The relationship between the semi-major axis, semi-minor axis, and the distance to the foci is given by:
\[ c^2 = a^2 - b^2 \]
- Substituting \(b = c\sqrt{3}\) into this equation:
\[ c^2 = a^2 - (c\sqrt{3})^2 \]
\[ c^2 = a^2 - 3c^2 \]
\[ 4c^2 = a^2 \]
\[ a = 2c \]

7. **Finding the Eccentricity**:
- The eccentricity \(e\) of the ellipse is given by:
\[ e = \frac{c}{a} \]
- Substituting \(a = 2c\):
\[ e = \frac{c}{2c} = \frac{1}{2} \]

### Conclusion:
The eccentricity of the ellipse is \( \frac{1}{2} \).