Find the ecentricity of ellipse `36x^(2) + 4y^(2) = 144`.

Answer

To find the eccentricity of the ellipse given by the equation \(36x^2 + 4y^2 = 144\), we can follow these steps:

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

### 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:
\[ a^2 = 4 \quad \text{and} \quad b^2 = 36 \]
Thus, we have:
\[ a = 2 \quad \text{and} \quad b = 6 \]

### Step 3: Calculate \(c\)
The value of \(c\) (the distance from the center to the foci) is calculated using the formula:
\[ c = \sqrt{b^2 - a^2} \]
Substituting the values we found:
\[ c = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2} \]

### Step 4: Calculate eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by the formula:
\[ e = \frac{c}{b} \]
Substituting the values of \(c\) and \(b\):
\[ e = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ \boxed{\frac{2\sqrt{2}}{3}} \]