The foci of an ellipse are `(-2,4)` and (2,1). The point `(1,(23)/(6))` is an extremity of the minor axis. What is the value of the eccentricity?

Options

A. `(9)/(13)`
B. `(3)/(sqrt(13))`
C. `(2)/(sqrt(13))`
D. `(4)/(13)`

Explanation

To find the eccentricity of the ellipse given the foci and an extremity of the minor axis, we can follow these steps:

### Step 1: Identify the coordinates of the foci
The foci of the ellipse are given as \( F_1(-2, 4) \) and \( F_2(2, 1) \).

### Step 2: Calculate the distance between the foci
The distance \( d \) between the foci can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of the foci:
\[ d = \sqrt{(2 - (-2))^2 + (1 - 4)^2} = \sqrt{(2 + 2)^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 3: Relate the distance between foci to eccentricity
The distance between the foci is given by \( 2ae \), where \( a \) is the semi-major axis and \( e \) is the eccentricity. Thus:
\[ 2ae = 5 \implies ae = \frac{5}{2} \]

### Step 4: Find the center of the ellipse
The center of the ellipse is the midpoint of the foci:
\[ \text{Center} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-2 + 2}{2}, \frac{4 + 1}{2} \right) = \left( 0, \frac{5}{2} \right) \]

### Step 5: Calculate the distance from the center to the extremity of the minor axis
The extremity of the minor axis is given as \( (1, \frac{23}{6}) \). We need to calculate the distance from the center \( (0, \frac{5}{2}) \) to this point:
\[ \text{Distance} = \sqrt{(1 - 0)^2 + \left(\frac{23}{6} - \frac{5}{2}\right)^2} \]
First, convert \( \frac{5}{2} \) to a fraction with a denominator of 6:
\[ \frac{5}{2} = \frac{15}{6} \]
Now calculate the distance:
\[ \text{Distance} = \sqrt{1^2 + \left(\frac{23}{6} - \frac{15}{6}\right)^2} = \sqrt{1 + \left(\frac{8}{6}\right)^2} = \sqrt{1 + \left(\frac{4}{3}\right)^2} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{9}{9} + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \]
This distance is equal to \( b \) (the semi-minor axis).

### Step 6: Use the relationship between \( a \), \( b \), and \( e \)
We know that:
\[ b^2 = a^2(1 - e^2) \]
Substituting \( b = \frac{5}{3} \):
\[ \left(\frac{5}{3}\right)^2 = a^2(1 - e^2) \implies \frac{25}{9} = a^2(1 - e^2) \]

### Step 7: Find \( a^2 \) using \( ae = \frac{5}{2} \)
From \( ae = \frac{5}{2} \), we can express \( a \) in terms of \( e \):
\[ a = \frac{5}{2e} \]
Substituting into the equation:
\[ \frac{25}{9} = \left(\frac{5}{2e}\right)^2(1 - e^2) \]
This simplifies to:
\[ \frac{25}{9} = \frac{25}{4e^2}(1 - e^2) \]
Cross-multiplying gives:
\[ 25 \cdot 4e^2 = 25 \cdot 9(1 - e^2) \implies 100e^2 = 225 - 225e^2 \]
Combining like terms:
\[ 325e^2 = 225 \implies e^2 = \frac{225}{325} = \frac{45}{65} = \frac{9}{13} \]
Thus,
\[ e = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}} \]

### Final Answer
The eccentricity \( e \) of the ellipse is:
\[ \boxed{\frac{3}{\sqrt{13}}} \]