What is the equation of the ellipse whose vertices are `(pm5, 0)` and foci are at `(pm, 4, 0)`?

Options

A. `(x^(2))/(25)+(y^(2))/(9)=1`
B. `(x^(2))/(16)+(y^(2))/(9)=1`
C. `(x^(2))/(25)+(y^(2))/(16)=1`
D. `(x^(2))/(9)+(y^(2))/(25)=1`

Answer

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

### Step 1: Identify the values of \( a \) and \( c \)
The vertices of the ellipse are given as \( (\pm 5, 0) \). This means the distance from the center to the vertices, \( a \), is 5. Therefore:
\[ a = 5 \]

The foci are given as \( (\pm 4, 0) \). This means the distance from the center to the foci, \( c \), is 4. Therefore:
\[ c = 4 \]

### Step 2: Use the relationship between \( a \), \( b \), and \( c \)
For an ellipse, the relationship between \( a \), \( b \), and \( c \) is given by the equation:
\[ c^2 = a^2 - b^2 \]
We already have \( a \) and \( c \):
\[ c^2 = 4^2 = 16 \]
\[ a^2 = 5^2 = 25 \]

Now substituting these values into the equation:
\[ 16 = 25 - b^2 \]

### Step 3: Solve for \( b^2 \)
Rearranging the equation gives:
\[ b^2 = 25 - 16 \]
\[ b^2 = 9 \]

### Step 4: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

### Final Answer
Thus, the equation of the ellipse is:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]