Find the eccentricity of the ellipse
(i) whose latus rectum is equal to half of its minor axis
(ii) whose latus rectum is equal to half of its major axis
(iii) if the major axis is three times the minor axis

Answer

To solve the problem of finding the eccentricity of the ellipse under the given conditions, we will go through each part step by step.

### Part (i): Latus Rectum is equal to half of its Minor Axis

1. **Understanding the Latus Rectum**: The formula for the latus rectum \( L \) of an ellipse is given by:
\[ L = \frac{2B^2}{A} \]
where \( A \) is the semi-major axis and \( B \) is the semi-minor axis.

2. **Minor Axis**: The length of the minor axis is \( 2B \). Therefore, half of the minor axis is:
\[ \text{Half of Minor Axis} = \frac{1}{2} \times 2B = B \]

3. **Setting up the equation**: According to the problem, the latus rectum is equal to half of the minor axis:
\[ \frac{2B^2}{A} = B \]

4. **Solving for A**: Rearranging the equation gives:
\[ 2B^2 = AB \implies A = 2B \]

5. **Finding Eccentricity**: The eccentricity \( e \) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} \]
Substituting \( A = 2B \):
\[ e = \sqrt{1 - \frac{B^2}{(2B)^2}} = \sqrt{1 - \frac{B^2}{4B^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Part (ii): Latus Rectum is equal to half of its Major Axis

1. **Setting up the equation**: The length of the major axis is \( 2A \), so half of the major axis is:
\[ \text{Half of Major Axis} = \frac{1}{2} \times 2A = A \]
According to the problem:
\[ \frac{2B^2}{A} = A \]

2. **Solving for A**: Rearranging the equation gives:
\[ 2B^2 = A^2 \implies A^2 = 2B^2 \]

3. **Finding Eccentricity**: Using the eccentricity formula:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{B^2}{2B^2}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Part (iii): Major Axis is three times the Minor Axis

1. **Setting up the relationship**: Given that the major axis is three times the minor axis:
\[ 2A = 3 \times 2B \implies A = 3B \]

2. **Finding Eccentricity**: Using the eccentricity formula:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{B^2}{(3B)^2}} = \sqrt{1 - \frac{B^2}{9B^2}} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

### Summary of Results
- (i) Eccentricity \( e = \frac{\sqrt{3}}{2} \)
- (ii) Eccentricity \( e = \frac{1}{\sqrt{2}} \)
- (iii) Eccentricity \( e = \frac{2\sqrt{2}}{3} \)