The eccentricity of the ellipse, which passes through the points (2, −2) and (−3,1) is :

Options

A. `(1)/(sqrt(5))`
B. `sqrt((2)/(5))`
C. `sqrt((3)/(5))`
D. `sqrt((4)/(5))`

Explanation

To find the eccentricity of the ellipse passing through the points (2, -2) and (-3, 1), we can follow these steps:

### Step 1: Write the general equation of the ellipse
The general equation of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

### Step 2: Substitute the first point (2, -2) into the equation
Substituting \(x = 2\) and \(y = -2\) into the ellipse equation:
\[ \frac{2^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} + \frac{4}{b^2} = 1 \]
Multiplying through by \(a^2b^2\) gives:
\[ 4b^2 + 4a^2 = a^2b^2 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (-3, 1) into the equation
Now, substituting \(x = -3\) and \(y = 1\):
\[ \frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} + \frac{1}{b^2} = 1 \]
Multiplying through by \(a^2b^2\) gives:
\[ 9b^2 + a^2 = a^2b^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
Now we have two equations:
1. \(4b^2 + 4a^2 = a^2b^2\)
2. \(9b^2 + a^2 = a^2b^2\)

From Equation 1, we can rearrange it to:
\[ a^2b^2 - 4a^2 - 4b^2 = 0 \]
From Equation 2, we can rearrange it to:
\[ a^2b^2 - 9b^2 - a^2 = 0 \]

### Step 5: Subtract Equation 1 from Equation 2
Subtracting Equation 1 from Equation 2 gives:
\[ (9b^2 + a^2) - (4b^2 + 4a^2) = 0 \]
This simplifies to:
\[ 5b^2 - 3a^2 = 0 \]
Thus, we have:
\[ \frac{b^2}{a^2} = \frac{3}{5} \]
This implies:
\[ b^2 = \frac{3}{5}a^2 \]

### Step 6: Find the eccentricity
The eccentricity \(e\) of an ellipse is given by:
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Substituting \(b^2 = \frac{3}{5}a^2\):
\[ e^2 = 1 - \frac{3}{5} = \frac{2}{5} \]
Thus:
\[ e = \sqrt{\frac{2}{5}} \]

### Final Answer
The eccentricity of the ellipse is:
\[ \boxed{\sqrt{\frac{2}{5}}} \]