Find the latus rectum, the eccentricity, and the coordinates of the foci, of the ellipses
` 5 x^(2) + 4y^(2) = 1 `

Answer

To solve the problem of finding the latus rectum, eccentricity, and coordinates of the foci of the ellipse given by the equation \(5x^2 + 4y^2 = 1\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
The standard form of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a^2\) and \(b^2\) are the denominators corresponding to the \(x^2\) and \(y^2\) terms respectively.

Starting with the given equation:
\[ 5x^2 + 4y^2 = 1 \]
we can rewrite it as:
\[ \frac{x^2}{\frac{1}{5}} + \frac{y^2}{\frac{1}{4}} = 1 \]
This implies:
\[ a^2 = \frac{1}{5} \quad \text{and} \quad b^2 = \frac{1}{4} \]

### Step 2: Identify \(a\) and \(b\)
Calculating \(a\) and \(b\):
\[ a = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \quad \text{and} \quad b = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 3: Determine the latus rectum
The formula for the latus rectum \(L\) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \(b^2\) and \(a\):
\[ L = \frac{2 \cdot \frac{1}{4}}{\frac{1}{\sqrt{5}}} = \frac{\frac{1}{2}}{\frac{1}{\sqrt{5}}} = \frac{1}{2} \cdot \sqrt{5} = \frac{\sqrt{5}}{2} \]

### Step 4: Calculate the eccentricity
The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 - \frac{\frac{1}{5}}{\frac{1}{4}}} = \sqrt{1 - \frac{4}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

### Step 5: Find the coordinates of the foci
The coordinates of the foci of an ellipse centered at the origin with a vertical major axis are given by:
\[ (0, \pm be) \]
Calculating \(be\):
\[ be = \frac{1}{2} \cdot \frac{1}{\sqrt{5}} = \frac{1}{2\sqrt{5}} \]
Thus, the coordinates of the foci are:
\[ (0, \pm \frac{1}{2\sqrt{5}}) \]

### Final Results
- Latus Rectum: \(\frac{\sqrt{5}}{2}\)
- Eccentricity: \(\frac{1}{\sqrt{5}}\)
- Coordinates of the Foci: \((0, \frac{1}{2\sqrt{5}})\) and \((0, -\frac{1}{2\sqrt{5}})\)