Find the latus rectum and eccentricity of the ellipse whose semi-axes are 5 and 4.

Answer

To find the latus rectum and eccentricity of the ellipse with semi-axes 5 and 4, we can follow these steps:

### Step 1: Identify the semi-major and semi-minor axes
Given:
- Semi-major axis \( A = 5 \)
- Semi-minor axis \( B = 4 \)

### Step 2: Write the equation of the ellipse
The standard form of the equation of an ellipse is:
\[ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \]
Substituting the values of \( A \) and \( B \):
\[ \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

### Step 3: Calculate the eccentricity
The formula for the eccentricity \( E \) of an ellipse is given by:
\[ E = \sqrt{1 - \frac{B^2}{A^2}} \]
Calculating \( B^2 \) and \( A^2 \):
- \( B^2 = 4^2 = 16 \)
- \( A^2 = 5^2 = 25 \)

Now substituting these values into the formula:
\[ E = \sqrt{1 - \frac{16}{25}} = \sqrt{1 - 0.64} = \sqrt{0.36} = \frac{3}{5} \]

### Step 4: Calculate the latus rectum
The formula for the latus rectum \( LR \) of an ellipse is:
\[ LR = \frac{2B^2}{A} \]
Substituting the values of \( B^2 \) and \( A \):
\[ LR = \frac{2 \cdot 16}{5} = \frac{32}{5} \]

### Final Results
- The eccentricity \( E \) of the ellipse is \( \frac{3}{5} \).
- The latus rectum \( LR \) of the ellipse is \( \frac{32}{5} \).

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