In the ellipse `(x^(2))/(6)+(y^(2))/(8)=1`, the value of eccentricity is

Options

A. `1/(2)`
B. `2/(3)2`
C. `1/(3)`
D. `3/(2)`

Answer

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

### Step 1: Identify the values of \(a^2\) and \(b^2\)
The standard form of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
From the given equation, we can identify:
- \(a^2 = 6\)
- \(b^2 = 8\)

### Step 2: Determine the values of \(a\) and \(b\)
To find \(a\) and \(b\), we take the square roots of \(a^2\) and \(b^2\):
\[ a = \sqrt{6} \]
\[ b = \sqrt{8} = 2\sqrt{2} \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Since \(b^2\) is greater than \(a^2\), we can use:
\[ e = \sqrt{1 - \frac{6}{8}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Final Answer
Thus, the eccentricity of the given ellipse is:
\[ e = \frac{1}{2} \]

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