Let S, S' be the focil and BB' be the minor axis of the ellipse `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1.` If `angle BSS' = theta`, then the eccentricity e of the ellipse is equal to

Options

A. `sin theta`
B. `cos theta`
C. `tan theta`
D. `cot theta`

Answer

To solve the problem, we need to find the eccentricity \( e \) of the ellipse given the angle \( \angle BSS' = \theta \).

### Step 1: Understand the Components of the Ellipse
The standard form of the ellipse is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. The foci \( S \) and \( S' \) are located at \( (c, 0) \) and \( (-c, 0) \) respectively, where \( c = \sqrt{a^2 - b^2} \).

**Hint:** Remember that the foci are located along the major axis, which is horizontal in this case.

### Step 2: Identify the Position of Points
The points \( B \) and \( B' \) are at the ends of the minor axis, which are located at \( (0, b) \) and \( (0, -b) \).

**Hint:** The minor axis is vertical, and its endpoints are directly above and below the center of the ellipse.

### Step 3: Analyze the Angle \( \angle BSS' \)
The angle \( \angle BSS' \) can be analyzed using the coordinates of the points.

- The coordinates of \( B \) are \( (0, b) \).
- The coordinates of \( S \) are \( (c, 0) \).
- The coordinates of \( S' \) are \( (-c, 0) \).

We can find the slopes of the lines \( BS \) and \( BS' \):

- Slope of line \( BS \) from \( B(0, b) \) to \( S(c, 0) \) is:

\[ \text{slope of } BS = \frac{0 - b}{c - 0} = -\frac{b}{c} \]

- Slope of line \( BS' \) from \( B(0, b) \) to \( S'(-c, 0) \) is:

\[ \text{slope of } BS' = \frac{0 - b}{-c - 0} = \frac{b}{c} \]

**Hint:** Use the slopes to find the angle between the two lines.

### Step 4: Use the Tangent Formula for Angle
The tangent of the angle \( \theta \) between the two lines can be expressed as:

\[ \tan(\theta) = \left| \frac{\text{slope of } BS - \text{slope of } BS'}{1 + \text{slope of } BS \cdot \text{slope of } BS'} \right| \]

Substituting the slopes:

\[ \tan(\theta) = \left| \frac{-\frac{b}{c} - \frac{b}{c}}{1 + \left(-\frac{b}{c}\right) \cdot \frac{b}{c}} \right| = \left| \frac{-\frac{2b}{c}}{1 - \frac{b^2}{c^2}} \right| \]

**Hint:** This formula helps relate the angle to the slopes of the lines.

### Step 5: Relate Eccentricity to the Angle
Recall that the eccentricity \( e \) of the ellipse is defined as:

\[ e = \frac{c}{a} \]

From the relationship of the ellipse, we know:

\[ c = \sqrt{a^2 - b^2} \]

Using the relationship between \( a \), \( b \), and \( e \):

\[ b = a \sqrt{1 - e^2} \]

Substituting \( b \) into the tangent equation gives a relationship involving \( e \) and \( \theta \).

**Hint:** Substitute \( b \) in terms of \( a \) and \( e \) to simplify the expression.

### Step 6: Solve for Eccentricity \( e \)
After substituting and simplifying, we can express \( e \) in terms of \( \tan(\theta) \):

\[ e = \frac{b \tan(\theta)}{\sqrt{a^2 - b^2}} \]

This leads us to the final expression for eccentricity based on the angle \( \theta \).

### Final Answer
Thus, the eccentricity \( e \) of the ellipse is given by:

\[ e = \frac{b \tan(\theta)}{a} \]