Let e the eccentricity of the ellipse passing through A(1, -1) and having foci at `F_1`(-2, 3) and `F_2`(5, 2), then `e^2` equals

Answer

To find the value of \( e^2 \) for the ellipse passing through the point \( A(1, -1) \) and having foci at \( F_1(-2, 3) \) and \( F_2(5, 2) \), we can follow these steps:

### Step 1: Calculate the distances from point A to the foci
We need to find the distances \( AF_1 \) and \( AF_2 \).

1. **Distance \( AF_1 \)**:
\[ AF_1 = \sqrt{(1 - (-2))^2 + (-1 - 3)^2} = \sqrt{(1 + 2)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

2. **Distance \( AF_2 \)**:
\[ AF_2 = \sqrt{(1 - 5)^2 + (-1 - 2)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 2: Use the definition of the ellipse
According to the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant and equals \( 2a \), where \( a \) is the semi-major axis.

From the distances calculated:
\[ AF_1 + AF_2 = 5 + 5 = 10 \]
Thus, we have:
\[ 2a = 10 \implies a = 5 \]

### Step 3: Calculate the distance between the foci
Next, we need to find the distance \( d \) between the foci \( F_1 \) and \( F_2 \):
\[ d = \sqrt{((-2) - 5)^2 + (3 - 2)^2} = \sqrt{(-7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} \]

### Step 4: Relate the distance between foci to the eccentricity
The distance between the foci is given by \( 2c \), where \( c \) is the distance from the center to each focus. We know:
\[ d = 2c \implies 2c = \sqrt{50} \implies c = \frac{\sqrt{50}}{2} = \frac{5\sqrt{2}}{2} \]

### Step 5: Use the relationship between \( a \), \( b \), and \( c \)
For an ellipse, the relationship between \( a \), \( b \), and \( c \) is given by:
\[ c^2 = a^2 - b^2 \]
We already have \( a = 5 \) and \( c = \frac{5\sqrt{2}}{2} \).

Calculating \( c^2 \):
\[ c^2 = \left(\frac{5\sqrt{2}}{2}\right)^2 = \frac{25 \cdot 2}{4} = \frac{50}{4} = 12.5 \]

Calculating \( a^2 \):
\[ a^2 = 5^2 = 25 \]

### Step 6: Solve for \( b^2 \)
Using the relationship:
\[ c^2 = a^2 - b^2 \implies 12.5 = 25 - b^2 \implies b^2 = 25 - 12.5 = 12.5 \]

### Step 7: Calculate \( e^2 \)
The eccentricity \( e \) is defined as:
\[ e = \frac{c}{a} \]
Thus:
\[ e^2 = \left(\frac{c}{a}\right)^2 = \frac{c^2}{a^2} = \frac{12.5}{25} = \frac{1}{2} \]

### Final Result
Therefore, the value of \( e^2 \) is:
\[ \boxed{\frac{1}{2}} \]