Find the eccentricity of the ellipse of minor axis 2b, if the line segment joining the foci subtend and angle `2 theta` at the upper vertex.

Answer

To find the eccentricity of the ellipse with a minor axis of length \(2b\) and the given angle \(2\theta\) subtended at the upper vertex by the line segment joining the foci, we can follow these steps:

### Step 1: Understand the properties of the ellipse
The standard form of an ellipse centered at the origin is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(2a\) is the length of the major axis and \(2b\) is the length of the minor axis. Here, we know that \(b = b\) and \(2b\) is given.

### Step 2: Identify the foci and the angle
The foci of the ellipse are located at \((c, 0)\) and \((-c, 0)\), where \(c = ae\) and \(e\) is the eccentricity. The angle \(2\theta\) subtended at the upper vertex (which is at \((0, b)\)) by the line segment joining the foci can be analyzed using trigonometry.

### Step 3: Use the tangent of the angle
From the upper vertex \((0, b)\), the distance to each focus is \(c\) and the vertical distance to the vertex is \(b\). The tangent of the angle \(\theta\) can be expressed as:
\[ \tan \theta = \frac{b}{c} \]
Thus, we can write:
\[ c = b \cot \theta \]

### Step 4: Relate \(c\), \(a\), and \(e\)
Since \(c = ae\), we can substitute \(c\) from the previous step:
\[ ae = b \cot \theta \]

### Step 5: Use the relationship between \(a\), \(b\), and \(e\)
We know the relationship:
\[ b^2 = a^2(1 - e^2) \]
Substituting \(c = ae\) into the equation gives:
\[ b^2 = a^2 - c^2 = a^2 - (b \cot \theta)^2 \]

### Step 6: Substitute \(c\) in terms of \(b\) and \(\theta\)
Substituting \(c\) into the equation:
\[ b^2 = a^2 - \frac{b^2}{\cos^2 \theta} \]
Rearranging gives:
\[ b^2 + \frac{b^2}{\cos^2 \theta} = a^2 \]
Factoring out \(b^2\):
\[ b^2\left(1 + \frac{1}{\cos^2 \theta}\right) = a^2 \]

### Step 7: Solve for \(e\)
Now, we can express \(e\) in terms of \(\theta\):
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Substituting \(a^2\) from the previous step into this equation gives:
\[ e^2 = 1 - \frac{b^2}{b^2\left(1 + \frac{1}{\cos^2 \theta}\right)} = 1 - \frac{1}{1 + \frac{1}{\cos^2 \theta}} = \frac{\frac{1}{\cos^2 \theta}}{1 + \frac{1}{\cos^2 \theta}} = \frac{1}{\cos^2 \theta + 1} \]

### Step 8: Final expression for \(e\)
Thus, the eccentricity \(e\) is given by:
\[ e = \sqrt{\frac{1}{\cos^2 \theta + 1}} = \frac{1}{\sqrt{1 + \tan^2 \theta}} = \sin \theta \]

### Conclusion
The eccentricity of the ellipse is:
\[ e = \sin \theta \]