Find the equation of the ellipse whose foci are at the points S(2, 0) and `S^(')` (-2, 0), and whose latus rectum is 6.

Answer

To find the equation of the ellipse with given foci and latus rectum, we can follow these steps:

### Step 1: Identify the foci and center
The foci of the ellipse are given as \( S(2, 0) \) and \( S'(-2, 0) \). The center of the ellipse, which is the midpoint of the line segment joining the foci, can be calculated as:
\[ \text{Center} = \left( \frac{2 + (-2)}{2}, \frac{0 + 0}{2} \right) = (0, 0) \]

### Step 2: Determine the distance \( c \)
The distance from the center to each focus is denoted as \( c \). From the foci, we can see:
\[ c = 2 \]

### Step 3: Use the latus rectum to find \( a \) and \( b \)
The latus rectum \( L \) of an ellipse is given as:
\[ L = \frac{2b^2}{a} \]
Given that the latus rectum is \( 6 \), we can set up the equation:
\[ \frac{2b^2}{a} = 6 \implies b^2 = 3a \]

### Step 4: Relate \( a \), \( b \), and \( c \)
We know the relationship between \( a \), \( b \), and \( c \) for an ellipse:
\[ c^2 = a^2 - b^2 \]
Substituting \( c = 2 \):
\[ 4 = a^2 - b^2 \]

### Step 5: Substitute \( b^2 \) into the equation
From the previous step, we have \( b^2 = 3a \). Substitute this into the equation:
\[ 4 = a^2 - 3a \]
Rearranging gives:
\[ a^2 - 3a - 4 = 0 \]

### Step 6: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \]
This gives us:
\[ a = \frac{8}{2} = 4 \quad \text{or} \quad a = \frac{-2}{2} = -1 \]
Since \( a \) must be positive, we have:
\[ a = 4 \]

### Step 7: Find \( b^2 \)
Now substitute \( a = 4 \) back into the equation for \( b^2 \):
\[ b^2 = 3a = 3 \times 4 = 12 \]

### Step 8: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin with the major axis along the x-axis is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = 16 \) and \( b^2 = 12 \):
\[ \frac{x^2}{16} + \frac{y^2}{12} = 1 \]

### Final Equation
Thus, the equation of the ellipse is:
\[ \frac{x^2}{16} + \frac{y^2}{12} = 1 \]