Find the eccentricity, distance between the foci, equation of the directrices, and the length and coordinates of the ends of the latus rectum of the ellipse `25x^(2)+16y^(2)=400`.

Answer

To solve the problem step by step, we will follow the outlined process to find the eccentricity, distance between the foci, equations of the directrices, and the length and coordinates of the ends of the latus rectum of the ellipse given by the equation \(25x^2 + 16y^2 = 400\).

### Step 1: Convert the given equation into standard form

The given equation is:
\[ 25x^2 + 16y^2 = 400 \]
To convert this into standard form, we divide both sides by 400:
\[ \frac{25x^2}{400} + \frac{16y^2}{400} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]
Now, we can identify \(a^2\) and \(b^2\):
\[ a^2 = 16 \quad \Rightarrow \quad a = 4 \]
\[ b^2 = 25 \quad \Rightarrow \quad b = 5 \]

### Step 2: Find the eccentricity

The eccentricity \(e\) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 3: Find the distance between the foci

The distance between the foci for an ellipse is given by:
\[ 2c \quad \text{where} \quad c = be \]
Calculating \(c\):
\[ c = b \cdot e = 5 \cdot \frac{3}{5} = 3 \]
Thus, the distance between the foci is:
\[ 2c = 2 \cdot 3 = 6 \]

### Step 4: Find the equations of the directrices

For an ellipse where \(a < b\), the equations of the directrices are given by:
\[ y = \pm \frac{b}{e} \]
Substituting the values of \(b\) and \(e\):
\[ y = \pm \frac{5}{\frac{3}{5}} = \pm \frac{25}{3} \]
Thus, the equations of the directrices are:
\[ y = \frac{25}{3} \quad \text{and} \quad y = -\frac{25}{3} \]

### Step 5: Find the coordinates of the ends of the latus rectum

The coordinates of the ends of the latus rectum can be found using:
\[ \left(\pm \frac{b^2}{a}, c\right) \]
Calculating:
\[ \frac{b^2}{a} = \frac{25}{4} \]
Thus, the coordinates of the ends of the latus rectum are:
\[ \left(\frac{25}{4}, 3\right) \quad \text{and} \quad \left(-\frac{25}{4}, 3\right) \]

### Step 6: Find the length of the latus rectum

The length of the latus rectum is given by:
\[ \frac{2b^2}{a} \]
Calculating:
\[ \text{Length} = \frac{2 \cdot 25}{4} = \frac{50}{4} = \frac{25}{2} \]

### Summary of Results
1. **Eccentricity**: \( \frac{3}{5} \)
2. **Distance between the foci**: \( 6 \)
3. **Equations of the directrices**: \( y = \frac{25}{3} \) and \( y = -\frac{25}{3} \)
4. **Coordinates of the ends of the latus rectum**: \( \left(\frac{25}{4}, 3\right) \) and \( \left(-\frac{25}{4}, 3\right) \)
5. **Length of the latus rectum**: \( \frac{25}{2} \)