Find the equation of the ellipse whose foci are `(0,pm1)` and eccentricity is `(1)/(2)`.

Answer

To find the equation of the ellipse whose foci are at (0, ±1) and eccentricity is \( \frac{1}{2} \), we can follow these steps:

### Step 1: Identify the form of the ellipse
Since the foci are located at (0, ±1), we can conclude that the ellipse is vertical. The standard form of a vertical ellipse is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \( b > a \).

### Step 2: Determine the value of \( b \)
The foci of a vertical ellipse are located at (0, ±c), where \( c = be \) (e is the eccentricity). Given that the foci are at (0, ±1), we have:

\[ c = 1 \]

The eccentricity \( e \) is given as \( \frac{1}{2} \). Therefore, we can express \( c \) in terms of \( b \):

\[ c = b \cdot e \]

Substituting the values we have:

\[ 1 = b \cdot \frac{1}{2} \]

Solving for \( b \):

\[ b = 2 \]

### Step 3: Use the relationship between \( a \), \( b \), and \( e \)
For an ellipse, the relationship between \( a \), \( b \), and \( e \) is given by:

\[ e^2 = 1 - \frac{a^2}{b^2} \]

Substituting \( e = \frac{1}{2} \) and \( b = 2 \):

\[ \left(\frac{1}{2}\right)^2 = 1 - \frac{a^2}{2^2} \]

This simplifies to:

\[ \frac{1}{4} = 1 - \frac{a^2}{4} \]

### Step 4: Solve for \( a^2 \)
Rearranging the equation gives:

\[ \frac{a^2}{4} = 1 - \frac{1}{4} \]

\[ \frac{a^2}{4} = \frac{3}{4} \]

Multiplying both sides by 4:

\[ a^2 = 3 \]

### Step 5: Write the equation of the ellipse
Now that we have \( a^2 = 3 \) and \( b^2 = 4 \), we can write the equation of the ellipse:

\[ \frac{x^2}{3} + \frac{y^2}{4} = 1 \]

### Final Answer
The equation of the required ellipse is:

\[ \frac{x^2}{3} + \frac{y^2}{4} = 1 \]

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