Find the equation of the ellipse whose foci are at (-2, 4) and (4, 4) and major and minor axes are 10 and 8 respectively. Also, find the eccentricity of the ellipse.

Answer

To find the equation of the ellipse whose foci are at (-2, 4) and (4, 4), and whose major and minor axes are 10 and 8 respectively, we can follow these steps:

### Step 1: Identify the given information
- Foci: F1 = (-2, 4) and F2 = (4, 4)
- Length of the major axis (2a) = 10
- Length of the minor axis (2b) = 8

### Step 2: Calculate the center of the ellipse
The center (h, k) of the ellipse can be found using the midpoint formula:
\[ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2} \]
Substituting the coordinates of the foci:
\[ h = \frac{-2 + 4}{2} = \frac{2}{2} = 1, \quad k = \frac{4 + 4}{2} = \frac{8}{2} = 4 \]
Thus, the center of the ellipse is (1, 4).

### Step 3: Determine the values of a and b
From the lengths of the axes:
\[ 2a = 10 \implies a = \frac{10}{2} = 5 \]
\[ 2b = 8 \implies b = \frac{8}{2} = 4 \]

### Step 4: Write the standard equation of the ellipse
Since the foci are horizontally aligned, the standard form of the equation of the ellipse is given by:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
Substituting the values of h, k, a, and b:
\[ \frac{(x - 1)^2}{5^2} + \frac{(y - 4)^2}{4^2} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{25} + \frac{(y - 4)^2}{16} = 1 \]

### Step 5: Find the eccentricity of the ellipse
The eccentricity (e) of an ellipse is given by the formula:
\[ e = \frac{c}{a} \]
where \( c \) is the distance from the center to each focus, calculated as:
\[ c = \sqrt{a^2 - b^2} \]
Calculating \( c \):
\[ c = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \]
Now substituting \( c \) and \( a \) into the eccentricity formula:
\[ e = \frac{c}{a} = \frac{3}{5} \]

### Final Answers
- The equation of the ellipse is:
\[ \frac{(x - 1)^2}{25} + \frac{(y - 4)^2}{16} = 1 \]
- The eccentricity of the ellipse is:
\[ e = \frac{3}{5} \]