If `alpha, beta` are the eccentric angles of the extermities of a focal chord of an ellipse, then eccentricity of the ellipse is

Options

A. `(sin alpha+ sin beta)/(sin(alpha+beta))`
B. `(cos alpha+ cos beta)/(cos(alpha+beta))`
C. `(sin(alpha+beta))/(sin(alpha+beta))`
D. none of these

Answer

To find the eccentricity \( e \) of the ellipse given that \( \alpha \) and \( \beta \) are the eccentric angles of the extremities of a focal chord, we can follow these steps:

### Step 1: Understand the Focal Chord
A focal chord of an ellipse is a line segment that passes through one of the foci of the ellipse and has its endpoints on the ellipse. The eccentric angles \( \alpha \) and \( \beta \) correspond to these endpoints.

### Step 2: Coordinates of the Points
The coordinates of the points on the ellipse corresponding to the angles \( \alpha \) and \( \beta \) are given by:
- For \( \alpha \): \( (a \cos \alpha, b \sin \alpha) \)
- For \( \beta \): \( (a \cos \beta, b \sin \beta) \)

Where \( a \) is the semi-major axis and \( b \) is the semi-minor axis of the ellipse.

### Step 3: Focus of the Ellipse
The focus of the ellipse is located at \( (ae, 0) \), where \( e \) is the eccentricity of the ellipse.

### Step 4: Collinearity Condition
The points \( (a \cos \alpha, b \sin \alpha) \), \( (a \cos \beta, b \sin \beta) \), and \( (ae, 0) \) are collinear. For three points to be collinear, the area of the triangle formed by these points must be zero.

### Step 5: Area of Triangle
The area \( A \) of the triangle formed by the points can be calculated using the determinant:
\[ A = \frac{1}{2} \left| \begin{array}{ccc} a \cos \alpha & b \sin \alpha & 1 \\ a \cos \beta & b \sin \beta & 1 \\ ae & 0 & 1 \end{array} \right| \]
Setting this area equal to zero gives us the condition for collinearity.

### Step 6: Calculate the Determinant
Calculating the determinant:
\[ \left| \begin{array}{ccc} a \cos \alpha & b \sin \alpha & 1 \\ a \cos \beta & b \sin \beta & 1 \\ ae & 0 & 1 \end{array} \right| = 0 \]
Expanding this determinant leads to:
\[ a \cos \alpha (b \sin \beta - 0) + b \sin \alpha (0 - ae) + 1(ae \sin \beta - a \cos \beta \cdot 0) = 0 \]
This simplifies to:
\[ ab \cos \alpha \sin \beta - abe \sin \alpha + ae \sin \beta = 0 \]

### Step 7: Rearranging the Equation
Rearranging the equation gives:
\[ ab \cos \alpha \sin \beta + ae \sin \beta = abe \sin \alpha \]
Factoring out \( \sin \beta \):
\[ \sin \beta (ab \cos \alpha + ae) = abe \sin \alpha \]

### Step 8: Solving for Eccentricity
Dividing both sides by \( ab \sin \beta \) gives:
\[ \frac{e}{\sin \beta} = \frac{\cos \alpha}{\sin \alpha} + \frac{e}{b} \]
This leads to the relationship involving \( e \), \( \alpha \), and \( \beta \).

### Final Step: Conclusion
After simplification, we find that:
\[ e = \frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} \]
This gives us the eccentricity of the ellipse in terms of the angles \( \alpha \) and \( \beta \).