The Foci of the ellipse `(x^(2))/(16)+(y^(2))/(25)=1` are

Options

A. `(0pm3)`
B. `(0pm4)`
C. `(0pm5)`
D. `(0pm2)`

Answer

To find the foci of the ellipse given by the equation \(\frac{x^2}{16} + \frac{y^2}{25} = 1\), we can follow these steps:

### Step 1: Identify the values of \(a\) and \(b\)
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
From the given equation, we can identify:
- \(a^2 = 16\) which implies \(a = \sqrt{16} = 4\)
- \(b^2 = 25\) which implies \(b = \sqrt{25} = 5\)

### Step 2: Determine the relationship between \(a\) and \(b\)
Since \(b > a\) (5 > 4), this indicates that the major axis of the ellipse is along the y-axis.

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse can be calculated using the formula:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 4: Find the coordinates of the foci
For an ellipse with the major axis along the y-axis, the foci are located at:
\[ (0, \pm be) \]
Substituting the values of \(b\) and \(e\):
\[ (0, \pm 5 \cdot \frac{3}{5}) = (0, \pm 3) \]

### Final Answer
Thus, the foci of the ellipse are:
\[ (0, 3) \text{ and } (0, -3) \]