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 that passes through the points (2, -2) and (-3, 1), we can follow these steps:

### Step 1: Write the standard equation of the ellipse
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis, \(b\) is the semi-minor axis, and the eccentricity \(e\) is defined as:
\[ e = \frac{c}{a} \]
with \(c^2 = a^2 - b^2\).

### Step 2: Substitute the first point (2, -2) into the ellipse equation
Substituting the coordinates of the first point (2, -2):
\[ \frac{(2)^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} + \frac{4}{b^2} = 1 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (-3, 1) into the ellipse equation
Now, substituting the coordinates of the second point (-3, 1):
\[ \frac{(-3)^2}{a^2} + \frac{(1)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} + \frac{1}{b^2} = 1 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \(\frac{4}{a^2} + \frac{4}{b^2} = 1\)
2. \(\frac{9}{a^2} + \frac{1}{b^2} = 1\)

From Equation 1, we can express \(\frac{4}{b^2}\) in terms of \(\frac{4}{a^2}\):
\[ \frac{4}{b^2} = 1 - \frac{4}{a^2} \quad \Rightarrow \quad \frac{1}{b^2} = \frac{1}{4} - \frac{1}{a^2} \]

Substituting this into Equation 2:
\[ \frac{9}{a^2} + \left(\frac{1}{4} - \frac{1}{a^2}\right) = 1 \]
This simplifies to:
\[ \frac{8}{a^2} + \frac{1}{4} = 1 \]
Multiplying through by \(4a^2\) gives:
\[ 32 + a^2 = 4a^2 \quad \Rightarrow \quad 3a^2 = 32 \quad \Rightarrow \quad a^2 = \frac{32}{3} \]

### Step 5: Find \(b^2\)
Substituting \(a^2\) back into Equation 1 to find \(b^2\):
\[ \frac{4}{\frac{32}{3}} + \frac{4}{b^2} = 1 \quad \Rightarrow \quad \frac{12}{32} + \frac{4}{b^2} = 1 \]
This simplifies to:
\[ \frac{3}{8} + \frac{4}{b^2} = 1 \quad \Rightarrow \quad \frac{4}{b^2} = \frac{5}{8} \quad \Rightarrow \quad b^2 = \frac{32}{5} \]

### Step 6: Calculate the eccentricity
Now we can find \(c^2\):
\[ c^2 = a^2 - b^2 = \frac{32}{3} - \frac{32}{5} \]
Finding a common denominator (15):
\[ c^2 = \frac{160}{15} - \frac{96}{15} = \frac{64}{15} \]

Now, we can find the eccentricity:
\[ e = \frac{c}{a} = \sqrt{\frac{64}{15}} \div \sqrt{\frac{32}{3}} = \sqrt{\frac{64}{15} \cdot \frac{3}{32}} = \sqrt{\frac{192}{480}} = \sqrt{\frac{2}{5}} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ e = \sqrt{\frac{2}{5}} \]

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