The equation of the ellipse whose one cocus is at `(4,0)` an whose eccentricity is `4/5,` is

Options

A. `(x ^(2))/(3 ^(2)) + (y^(2))/(5 ^(2)) =1`
B. ` (x ^(2))/(5 ^(2)) + (y^(2))/(3 ^(2)) =1`
C. `(x ^(2))/(5 ^(2)) + (y ^(2))/(4 ^(2)) =1`
D. `(x ^(2))/( 4 ^(2)) + (y ^(2))/(5 ^(2)) =1`

Answer

To find the equation of the ellipse whose one focus is at (4,0) and whose eccentricity is \( \frac{4}{5} \), we can follow these steps:

### Step 1: Identify the parameters of the ellipse
Given:
- One focus \( F = (4, 0) \)
- Eccentricity \( e = \frac{4}{5} \)

### Step 2: Determine the value of \( a \)
The focus of an ellipse is given by the coordinates \( (ae, 0) \). Since one focus is at \( (4, 0) \), we can set up the equation:
\[ ae = 4 \]
Substituting the value of \( e \):
\[ a \cdot \frac{4}{5} = 4 \]
To find \( a \), we can solve for it:
\[ a = 4 \cdot \frac{5}{4} = 5 \]

### Step 3: Use the eccentricity to find \( b \)
The relationship between \( a \), \( b \), and \( e \) for an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \( e = \frac{4}{5} \) and \( a = 5 \):
\[ \frac{4}{5} = \sqrt{1 - \frac{b^2}{5^2}} \]
Squaring both sides:
\[ \left(\frac{4}{5}\right)^2 = 1 - \frac{b^2}{25} \]
\[ \frac{16}{25} = 1 - \frac{b^2}{25} \]
Rearranging gives:
\[ \frac{b^2}{25} = 1 - \frac{16}{25} \]
\[ \frac{b^2}{25} = \frac{9}{25} \]
Multiplying both sides by 25:
\[ b^2 = 9 \]
Taking the square root gives:
\[ b = 3 \]

### Step 4: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( a = 5 \) and \( b = 3 \):
\[ \frac{x^2}{5^2} + \frac{y^2}{3^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]