S and T are the foci of the ellipse `(x^2/a^2)+(y^2/b^2)=1` and B is an end of the minor axis. If STB is equilateral triangle, then eccentricity of the ellipse is

Options

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

Answer

To solve the problem, we need to find the eccentricity of the ellipse given that the points S, T (the foci), and B (an endpoint of the minor axis) form an equilateral triangle.

### Step-by-Step Solution:

1. **Identify the foci and endpoint of the minor axis**:
The foci of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) are located at \(S(ae, 0)\) and \(T(-ae, 0)\), where \(e\) is the eccentricity of the ellipse. The endpoint of the minor axis is at \(B(0, b)\).

2. **Understand the properties of the equilateral triangle**:
Since \(STB\) is an equilateral triangle, all angles are \(60^\circ\). Therefore, the angles \(OSB\), \(OTB\), and \(OST\) are all \(60^\circ\).

3. **Use trigonometric ratios**:
In triangle \(OSB\), we can use the tangent of angle \(OSB\):
\[ \tan(60^\circ) = \sqrt{3} = \frac{OB}{OS} \]
Here, \(OB = b\) (the length of the minor axis) and \(OS = ae\) (the distance from the center to the focus).

4. **Set up the equation**:
From the tangent ratio, we can write:
\[ \sqrt{3} = \frac{b}{ae} \]
Rearranging gives:
\[ b = ae\sqrt{3} \]

5. **Use the relationship between \(a\), \(b\), and \(e\)**:
We know that the relationship between \(a\), \(b\), and \(e\) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \(b = ae\sqrt{3}\) into this equation:
\[ e = \sqrt{1 - \frac{(ae\sqrt{3})^2}{a^2}} = \sqrt{1 - 3e^2} \]

6. **Square both sides to eliminate the square root**:
\[ e^2 = 1 - 3e^2 \]
Rearranging gives:
\[ 4e^2 = 1 \]

7. **Solve for \(e\)**:
\[ e^2 = \frac{1}{4} \implies e = \frac{1}{2} \]

Thus, the eccentricity of the ellipse is \( \frac{1}{2} \).