Find the eccentricity of the ellipse whose latus rectum is (i) half its major axis, (ii) half its minor axis.

Answer

To find the eccentricity of the ellipse whose latus rectum is half of its major axis and half of its minor axis, we will follow these steps:

### Step 1: Understand the Definitions
The latus rectum of an ellipse is a line segment that passes through a focus and is perpendicular to the major axis. The length of the latus rectum (L) can be expressed in terms of the semi-major axis (a) and the semi-minor axis (b) as:

\[ L = \frac{2b^2}{a} \]

### Step 2: Case (i) - Latus Rectum is Half of the Major Axis
The length of the major axis is \(2a\). If the latus rectum is half of the major axis, we have:

\[ L = \frac{1}{2} \times 2a = a \]

Setting the two expressions for the latus rectum equal gives:

\[ \frac{2b^2}{a} = a \]

### Step 3: Solve for b^2
Multiplying both sides by \(a\) gives:

\[ 2b^2 = a^2 \]

Dividing both sides by 2:

\[ b^2 = \frac{a^2}{2} \]

### Step 4: Use the Eccentricity Formula
The eccentricity \(e\) of an ellipse is given by the formula:

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

Substituting \(b^2 = \frac{a^2}{2}\) into the formula:

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

### Step 5: Case (ii) - Latus Rectum is Half of the Minor Axis
The length of the minor axis is \(2b\). If the latus rectum is half of the minor axis, we have:

\[ L = \frac{1}{2} \times 2b = b \]

Setting the two expressions for the latus rectum equal gives:

\[ \frac{2b^2}{a} = b \]

### Step 6: Solve for b
Multiplying both sides by \(a\) gives:

\[ 2b^2 = ab \]

Rearranging gives:

\[ 2b^2 - ab = 0 \]

Factoring out \(b\):

\[ b(2b - a) = 0 \]

Since \(b \neq 0\), we have:

\[ 2b = a \implies b = \frac{a}{2} \]

### Step 7: Use the Eccentricity Formula Again
Substituting \(b = \frac{a}{2}\) into the eccentricity formula:

\[ b^2 = \left(\frac{a}{2}\right)^2 = \frac{a^2}{4} \]

Now substituting into the eccentricity formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{\frac{a^2}{4}}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Final Answers
1. For the case where the latus rectum is half of the major axis, the eccentricity \(e = \frac{1}{\sqrt{2}}\).
2. For the case where the latus rectum is half of the minor axis, the eccentricity \(e = \frac{\sqrt{3}}{2}\).