The foci of the ellipse `(x^(2))/(4) +(y^(2))/(25) = 1 ` are :

Options

A. `(pm sqrt(5) , 0 ) `
B. `(-5, 0 ) `
C. `(0 , -5) `
D. none of these.

Answer

To find the foci of the ellipse given by the equation \(\frac{x^2}{4} + \frac{y^2}{25} = 1\), we can follow these steps:

### Step 1: Identify the values of \(a\) and \(b\)
The given equation of the ellipse can be compared to the standard form of the ellipse:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

From the given equation, we can identify:
- \(a^2 = 4\) which gives \(a = \sqrt{4} = 2\)
- \(b^2 = 25\) which gives \(b = \sqrt{25} = 5\)

### Step 2: Determine the relationship between \(a\) and \(b\)
Since \(b > a\) (5 > 2), this indicates that the major axis is along the y-axis.

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of an ellipse is given by the formula:

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

Substituting the values of \(a\) and \(b\):

\[ e = \sqrt{1 - \frac{4}{25}} = \sqrt{1 - 0.16} = \sqrt{0.84} = \frac{\sqrt{21}}{5} \]

### Step 4: Find the coordinates of the foci
The foci of the ellipse are located at \((0, \pm be)\). We need to calculate \(be\):

\[ be = b \cdot e = 5 \cdot \frac{\sqrt{21}}{5} = \sqrt{21} \]

Thus, the coordinates of the foci are:

\[ (0, \pm \sqrt{21}) \]

### Final Answer
The foci of the ellipse \(\frac{x^2}{4} + \frac{y^2}{25} = 1\) are at the points:

\[ (0, \sqrt{21}) \quad \text{and} \quad (0, -\sqrt{21}) \]

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