Find the eccentricity of the ellipse
`" "(x-3)^(2)/8+(y-4)^(2)/6=1`.

Answer

To find the eccentricity of the ellipse given by the equation \((x-3)^{2}/8 + (y-4)^{2}/6 = 1\), we can follow these steps:

### Step 1: Identify the standard form of the ellipse
The standard form of the equation of an ellipse is given by:
\[ \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \]
where \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.

### Step 2: Compare with the given equation
The given equation is:
\[ \frac{(x-3)^{2}}{8} + \frac{(y-4)^{2}}{6} = 1 \]
From this, we can identify:
- \(h = 3\)
- \(k = 4\)
- \(a^{2} = 8\)
- \(b^{2} = 6\)

### Step 3: Determine \(a\) and \(b\)
Since \(a^{2} = 8\) and \(b^{2} = 6\), we can find:
- \(a = \sqrt{8} = 2\sqrt{2}\)
- \(b = \sqrt{6}\)

### Step 4: Identify the relationship between \(a\) and \(b\)
Since \(a^{2} > b^{2}\) (because \(8 > 6\)), the major axis is along the x-axis.

### Step 5: Calculate the eccentricity
The formula for eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \]
Substituting the values of \(a^{2}\) and \(b^{2}\):
\[ e = \sqrt{1 - \frac{6}{8}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Final Result
Thus, the eccentricity of the ellipse is:
\[ \boxed{\frac{1}{2}} \]