Distance between foci is 4 and distance between directrices is 5

Options

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

Answer

To solve the problem regarding the ellipse with given distances between the foci and directrices, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The distance between the foci (2ae) is given as 4.
- The distance between the directrices (2a/e) is given as 5.

2. **Set Up the Equations**:
- From the distance between the foci:
\[ 2ae = 4 \implies ae = 2 \quad \text{(Equation 1)} \]
- From the distance between the directrices:
\[ \frac{2a}{e} = 5 \implies 2a = 5e \quad \text{(Equation 2)} \]

3. **Express 'a' in Terms of 'e'**:
- From Equation 1, we can express 'a':
\[ a = \frac{2}{e} \quad \text{(Substituting into Equation 2)} \]

4. **Substitute 'a' into Equation 2**:
- Substitute \( a = \frac{2}{e} \) into Equation 2:
\[ 2\left(\frac{2}{e}\right) = 5e \]
\[ \frac{4}{e} = 5e \]

5. **Solve for 'e'**:
- Multiply both sides by \( e \) to eliminate the fraction:
\[ 4 = 5e^2 \]
\[ e^2 = \frac{4}{5} \implies e = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]

6. **Find 'a'**:
- Substitute \( e \) back into \( a = \frac{2}{e} \):
\[ a = \frac{2}{\frac{2}{\sqrt{5}}} = \sqrt{5} \]

7. **Find 'b'**:
- Use the relationship \( e^2 = 1 - \frac{b^2}{a^2} \):
\[ \frac{4}{5} = 1 - \frac{b^2}{(\sqrt{5})^2} \]
\[ \frac{4}{5} = 1 - \frac{b^2}{5} \]
\[ \frac{b^2}{5} = 1 - \frac{4}{5} = \frac{1}{5} \]
\[ b^2 = 1 \]

8. **Write the Equation of the Ellipse**:
- The standard form of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
- Substituting \( a^2 = 5 \) and \( b^2 = 1 \):
\[ \frac{x^2}{5} + \frac{y^2}{1} = 1 \]
- Multiplying through by 5 gives:
\[ 5x^2 + 5y^2 = 5 \]

### Final Equation:
The equation of the ellipse is:
\[ 5x^2 + 5y^2 = 5 \quad \text{or} \quad x^2 + y^2 = 1 \]