What is the equation of the ellipse having foci (±2, 0) and the eccentricity `(1)/(4)`?

Options

A. `(x^(2))/(64)+(y^(2))/(60)=1`
B. `(x^(2))/(60)+(y^(2))/(64)=1`
C. `(x^(2))/(20)+(y^(2))/(24)=1`
D. `(x^(2))/(24)+(y^(2))/(20)=1`

Answer

To find the equation of the ellipse with foci at (±2, 0) and an eccentricity of \( \frac{1}{4} \), we can follow these steps:

### Step 1: Identify the values of \( c \) and \( e \)
The foci of the ellipse are given as (±2, 0). The distance from the center to each focus is denoted as \( c \). Therefore, we have:
\[ c = 2 \]

The eccentricity \( e \) is given as:
\[ e = \frac{1}{4} \]

### Step 2: Use the relationship between \( a \), \( b \), and \( c \)
For an ellipse, the relationship between the semi-major axis \( a \), semi-minor axis \( b \), and the distance to the foci \( c \) is given by:
\[ e = \frac{c}{a} \]
From this, we can find \( a \):
\[ a = \frac{c}{e} = \frac{2}{\frac{1}{4}} = 2 \times 4 = 8 \]

### Step 3: Calculate \( b \) using the relationship \( c^2 = a^2 - b^2 \)
Now we can use the relationship:
\[ c^2 = a^2 - b^2 \]
Substituting the known values:
\[ 2^2 = 8^2 - b^2 \]
\[ 4 = 64 - b^2 \]
Rearranging gives:
\[ b^2 = 64 - 4 = 60 \]

### Step 4: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin with a horizontal major axis is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ \frac{x^2}{64} + \frac{y^2}{60} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{64} + \frac{y^2}{60} = 1 \]

---