Find the equation of the ellipse whose foci are `(0,pm3)` and eccentricity is `(3)/(5)`.

Answer

To find the equation of the ellipse whose foci are at (0, ±3) and eccentricity is \( \frac{3}{5} \), we can follow these steps:

### Step 1: Identify the center and orientation of the ellipse
The foci are given as (0, ±3), which indicates that the ellipse is vertically oriented and centered at the origin (0, 0).

### Step 2: Determine the values of \( c \) and \( e \)
The distance from the center to each focus is denoted as \( c \). Here, \( c = 3 \) (the distance from the center to the foci at (0, ±3)). The eccentricity \( e \) is given as \( \frac{3}{5} \).

### Step 3: Use the relationship between \( c \), \( a \), and \( b \)
For an ellipse, the relationship between the semi-major axis \( a \), semi-minor axis \( b \), and the distance to the foci \( c \) is given by:
\[ c = ae \]
Substituting the known values:
\[ 3 = a \cdot \frac{3}{5} \]
To find \( a \), we can rearrange this equation:
\[ a = \frac{3 \cdot 5}{3} = 5 \]

### Step 4: Use the relationship between \( a \), \( b \), and \( c \)
We also know another relationship:
\[ c^2 = a^2 - b^2 \]
Substituting the known values:
\[ 3^2 = 5^2 - b^2 \]
This simplifies to:
\[ 9 = 25 - b^2 \]
Rearranging gives:
\[ b^2 = 25 - 9 = 16 \]
Thus, \( b = 4 \).

### Step 5: Write the equation of the ellipse
Since the ellipse is vertically oriented, the standard form of the equation is:
\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]
Substituting \( a^2 = 25 \) and \( b^2 = 16 \):
\[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]