Find the eccentricity, the coordinates of the foci, and the length of the latus rectum of the ellipse `2x^(2)+3y^(2)=1`.

Answer

To solve the problem step by step, we will find the eccentricity, the coordinates of the foci, and the length of the latus rectum of the ellipse given by the equation \(2x^2 + 3y^2 = 1\).

### Step 1: Rewrite the equation in standard form
We start with the equation of the ellipse:
\[ 2x^2 + 3y^2 = 1 \]
To rewrite it in standard form, we divide each term by 1:
\[ \frac{x^2}{\frac{1}{2}} + \frac{y^2}{\frac{1}{3}} = 1 \]
This can be expressed as:
\[ \frac{x^2}{\frac{1}{2}} + \frac{y^2}{\frac{1}{3}} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form, we can identify:
\[ a^2 = \frac{1}{2}, \quad b^2 = \frac{1}{3} \]
Thus, we find:
\[ a = \frac{1}{\sqrt{2}}, \quad b = \frac{1}{\sqrt{3}} \]

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

### Step 4: Find the coordinates of the foci
For an ellipse where \(a > b\), the coordinates of the foci are given by:
\[ (\pm ae, 0) \]
Substituting the values of \(a\) and \(e\):
\[ \text{Coordinates of foci} = \left(\pm \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}}, 0\right) = \left(\pm \frac{1}{\sqrt{6}}, 0\right) \]

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

### Final Results
1. **Eccentricity \(e\)**: \(\frac{1}{\sqrt{3}}\)
2. **Coordinates of the foci**: \(\left(\frac{1}{\sqrt{6}}, 0\right), \left(-\frac{1}{\sqrt{6}}, 0\right)\)
3. **Length of the latus rectum**: \(\frac{2\sqrt{2}}{3}\)