If (-4, 3) and (8, 3) are the vertices of an ellipse whose eccentricity is 5/6 then the equation of the ellipse is

Options

A. `((x - 2)^(2))/(11) + ((y - 3)^(2))/(36) = 1`
B. `((x - 2)^(2))/(36) + ((y - 3)^(2))/(11) = 1`
C. `((x - 3)^(2))/(36) + ((y - 2)^(2))/(11) = 1`
D. none of these

Answer

To find the equation of the ellipse with vertices at (-4, 3) and (8, 3) and an eccentricity of \( \frac{5}{6} \), we can follow these steps:

### Step 1: Identify the center of the ellipse
The center of the ellipse is the midpoint of the line segment joining the two vertices.

\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-4 + 8}{2}, \frac{3 + 3}{2} \right) = \left( \frac{4}{2}, 3 \right) = (2, 3) \]

### Step 2: Determine the length of the semi-major axis (a)
The distance from the center to either vertex gives us the length of the semi-major axis \( a \).

\[ a = \text{Distance from center to vertex} = |8 - 2| = 6 \]

### Step 3: Use the eccentricity to find the semi-minor axis (b)
The eccentricity \( e \) is given by the formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

We know \( e = \frac{5}{6} \) and \( a = 6 \). Plugging in these values:

\[ \left( \frac{5}{6} \right)^2 = 1 - \frac{b^2}{6^2} \]

Calculating \( a^2 \):

\[ \frac{25}{36} = 1 - \frac{b^2}{36} \]

Rearranging gives:

\[ \frac{b^2}{36} = 1 - \frac{25}{36} = \frac{11}{36} \]

Thus,

\[ b^2 = 11 \]

### Step 4: Write the equation of the ellipse
Since the major axis is horizontal (parallel to the x-axis), the standard form of the equation of the ellipse is:

\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]

Substituting \( h = 2 \), \( k = 3 \), \( a^2 = 36 \), and \( b^2 = 11 \):

\[ \frac{(x - 2)^2}{36} + \frac{(y - 3)^2}{11} = 1 \]

### Final Answer
The equation of the ellipse is:

\[ \frac{(x - 2)^2}{36} + \frac{(y - 3)^2}{11} = 1 \]