Eccentricity of ellipse `2(x-y+1)^(2)+3(x+y+2)^(2)=5` is

Options

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

Explanation

To find the eccentricity of the ellipse given by the equation \( 2(x-y+1)^{2}+3(x+y+2)^{2}=5 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form

We start with the given equation:
\[ 2(x-y+1)^{2}+3(x+y+2)^{2}=5 \]
We can divide the entire equation by 5 to simplify it:
\[ \frac{2(x-y+1)^{2}}{5} + \frac{3(x+y+2)^{2}}{5} = 1 \]

### Step 2: Identify the coefficients

Now, we can rewrite the equation in the form:
\[ \frac{(x-y+1)^{2}}{a^{2}} + \frac{(x+y+2)^{2}}{b^{2}} = 1 \]
where:
\[ a^{2} = \frac{5}{2} \quad \text{and} \quad b^{2} = \frac{5}{3} \]

### Step 3: Determine \(a\) and \(b\)

From the above, we have:
\[ a = \sqrt{\frac{5}{2}} \quad \text{and} \quad b = \sqrt{\frac{5}{3}} \]

### Step 4: Calculate the eccentricity

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

### Step 5: Simplify the expression

Now, simplifying the fraction:
\[ e = \sqrt{1 - \frac{5 \cdot 2}{5 \cdot 3}} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer

Thus, the eccentricity of the ellipse is:
\[ \frac{1}{\sqrt{3}} \]

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