Find the equation of the ellipse whose eccentricity is `1/2` and whose foci are at the points `(pm2, 0)`.

Answer

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

### Step 1: Identify the given values
We know:
- Eccentricity (e) = 1/2
- Foci = (±2, 0)

### Step 2: Determine the value of 'c'
The foci of the ellipse are located at (±c, 0). Here, c = 2 (from the foci).

### Step 3: Relate 'a', 'b', and 'c' using the eccentricity
The relationship between the semi-major axis (a), semi-minor axis (b), and the eccentricity (e) is given by the formula:
\[ e = \frac{c}{a} \]

Substituting the known values:
\[ \frac{1}{2} = \frac{2}{a} \]

### Step 4: Solve for 'a'
Cross-multiplying gives:
\[ 1 \cdot a = 2 \cdot 2 \]
\[ a = 4 \]

### Step 5: Calculate 'a^2'
Now, we calculate \( a^2 \):
\[ a^2 = 4^2 = 16 \]

### Step 6: Use the relationship to find 'b'
We also know the relationship:
\[ c^2 = a^2 - b^2 \]
Substituting the known values:
\[ 2^2 = 16 - b^2 \]
\[ 4 = 16 - b^2 \]

### Step 7: Solve for 'b^2'
Rearranging gives:
\[ b^2 = 16 - 4 \]
\[ b^2 = 12 \]

### 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 the values of \( a^2 \) and \( b^2 \):
\[ \frac{x^2}{16} + \frac{y^2}{12} = 1 \]

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

---