If the foci of an ellipse are `(pm sqrt5,0) and ` its exccentricity is `(sqrt5)/(3), ` then the equation of the ellipse is

Options

A. `9x ^(2) + 4y ^(2) =36`
B. `4x ^(2) + 9y ^(2) =36`
C. `36 x ^(2) + 9y ^(2) =4`
D. `4x ^(2) + 36 y ^(2) =4`

Answer

To find the equation of the ellipse given its foci and eccentricity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the foci and eccentricity**:
The foci of the ellipse are given as \((\pm \sqrt{5}, 0)\) and the eccentricity \(e\) is given as \(\frac{\sqrt{5}}{3}\).

2. **Determine the value of \(a\)**:
The foci of an ellipse are given by the formula \((\pm ae, 0)\). From the foci, we can see that:
\[ ae = \sqrt{5} \]
We know \(e = \frac{\sqrt{5}}{3}\). Substituting this into the equation gives:
\[ a \cdot \frac{\sqrt{5}}{3} = \sqrt{5} \]
To find \(a\), we can rearrange this:
\[ a = \sqrt{5} \cdot \frac{3}{\sqrt{5}} = 3 \]

3. **Use the relationship between \(a\), \(b\), and \(e\)**:
The eccentricity of an ellipse is also defined by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Squaring both sides gives:
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Substituting \(e = \frac{\sqrt{5}}{3}\):
\[ \left(\frac{\sqrt{5}}{3}\right)^2 = 1 - \frac{b^2}{3^2} \]
This simplifies to:
\[ \frac{5}{9} = 1 - \frac{b^2}{9} \]

4. **Solve for \(b^2\)**:
Rearranging the equation gives:
\[ \frac{b^2}{9} = 1 - \frac{5}{9} \]
\[ \frac{b^2}{9} = \frac{4}{9} \]
Multiplying both sides by 9:
\[ b^2 = 4 \]

5. **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 \(a^2 = 9\) and \(b^2 = 4\):
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]

### Final Equation:
The equation of the ellipse is:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]