Find the equation to the ellipse with axes as the axes of coordinates.
foci are `(pm 4,0)` and `e=1/3`,

Answer

To find the equation of the ellipse with foci at \((\pm 4, 0)\) and eccentricity \(e = \frac{1}{3}\), we can follow these steps:

### Step 1: Identify the values of \(c\) and \(e\)
The foci of the ellipse are given as \((\pm 4, 0)\). The distance from the center to each focus is denoted as \(c\). Therefore, we have:
\[ c = 4 \]

The eccentricity \(e\) is given as:
\[ e = \frac{1}{3} \]

### Step 2: Relate \(a\), \(b\), and \(c\)
For an ellipse, the relationship between \(a\), \(b\), and \(c\) is given by the equation:
\[ c = ae \]
Substituting the known values:
\[ 4 = a \cdot \frac{1}{3} \]
To find \(a\), we can rearrange this equation:
\[ a = 4 \cdot 3 = 12 \]

### Step 3: Calculate \(b\)
We can use the relationship between \(a\), \(b\), and \(c\):
\[ c^2 = a^2 - b^2 \]
Substituting the values we have:
\[ 4^2 = 12^2 - b^2 \]
Calculating \(4^2\) and \(12^2\):
\[ 16 = 144 - b^2 \]
Rearranging gives:
\[ b^2 = 144 - 16 = 128 \]

### Step 4: 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 \(b^2\):
\[ \frac{x^2}{12^2} + \frac{y^2}{128} = 1 \]
This simplifies to:
\[ \frac{x^2}{144} + \frac{y^2}{128} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{144} + \frac{y^2}{128} = 1 \]

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