The eccentricity of the ellipse `x^(2) + 2y^(2) =6` is

Options

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

Answer

To find the eccentricity of the ellipse given by the equation \( x^2 + 2y^2 = 6 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
Start with the given equation:
\[ x^2 + 2y^2 = 6 \]
To convert it to standard form, divide the entire equation by 6:
\[ \frac{x^2}{6} + \frac{2y^2}{6} = 1 \]
This simplifies to:
\[ \frac{x^2}{6} + \frac{y^2}{3} = 1 \]

### 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:
\[ A^2 = 6 \quad \text{and} \quad B^2 = 3 \]
Thus, we have:
\[ A = \sqrt{6} \quad \text{and} \quad B = \sqrt{3} \]

### Step 3: Determine the eccentricity
The formula for the eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} \quad \text{if } A > B \]
Since \(A = \sqrt{6}\) and \(B = \sqrt{3}\), we have \(A > B\). Therefore, we can use the formula:
\[ e = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{3}{6}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
The eccentricity of the ellipse is:
\[ e = \frac{1}{\sqrt{2}} \]

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