Find the ecentricity and foci of the ellipse `4x^(2) + 9y^(2) = 144`

Answer

To find the eccentricity and foci of the ellipse given by the equation \(4x^2 + 9y^2 = 144\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ 4x^2 + 9y^2 = 144 \]
To rewrite this in standard form, we divide every term by 144:
\[ \frac{4x^2}{144} + \frac{9y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{36} + \frac{y^2}{16} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = 36 \quad \text{and} \quad b^2 = 16 \]
Thus, we find:
\[ a = \sqrt{36} = 6 \quad \text{and} \quad b = \sqrt{16} = 4 \]

### Step 3: Calculate \(c\)
The value of \(c\) (the distance from the center to each focus) is given by the formula:
\[ c = \sqrt{a^2 - b^2} \]
Substituting the values we found:
\[ c = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \]

### Step 4: Find the coordinates of the foci
For an ellipse centered at the origin with the major axis along the x-axis, the coordinates of the foci are given by:
\[ (\pm c, 0) \]
Substituting the value of \(c\):
\[ \text{Foci} = (\pm 2\sqrt{5}, 0) \]

### Step 5: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by:
\[ e = \frac{c}{a} \]
Substituting the values we found:
\[ e = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \]

### Final Results
- The eccentricity \(e\) of the ellipse is \(\frac{\sqrt{5}}{3}\).
- The coordinates of the foci are \((-2\sqrt{5}, 0)\) and \((2\sqrt{5}, 0)\).