The equation of ellipse whose vertices are `(pm 5, 0)` and foci are `(pm 4, 0)` is (i) `(x^(2))/(16)+(y^(2))/(9)=1` (ii) `(x^(2))/(9)+(y^(2))/(16)=1` (iii) `(x^(2))/(25)+(y^(2))/(9)=1` (iv) `(x^(2))/(9)+(y^(2))/(25)=1`

Options

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

Answer

To find the equation of the ellipse whose vertices are at \((\pm 5, 0)\) and foci are at \((\pm 4, 0)\), we can follow these steps:

### Step 1: Identify the values of \(a\) and \(c\)
The vertices of the ellipse are given as \((\pm a, 0)\), which means:
- \(a = 5\)

The foci of the ellipse are given as \((\pm c, 0)\), which means:
- \(c = 4\)

### Step 2: Use the relationship between \(a\), \(b\), and \(c\)
For an ellipse, the relationship between \(a\), \(b\), and \(c\) is given by:
\[ c^2 = a^2 - b^2 \]

### Step 3: Calculate \(b^2\)
Substituting the known values of \(a\) and \(c\):
\[ c^2 = 4^2 = 16 \]
\[ a^2 = 5^2 = 25 \]
Now, substituting these into the relationship:
\[ 16 = 25 - b^2 \]
Rearranging gives:
\[ b^2 = 25 - 16 = 9 \]

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

### Step 5: Identify the correct option
Now we can compare this equation with the given options:
- (i) \(\frac{x^2}{16} + \frac{y^2}{9} = 1\)
- (ii) \(\frac{x^2}{9} + \frac{y^2}{16} = 1\)
- (iii) \(\frac{x^2}{25} + \frac{y^2}{9} = 1\) **(Correct)**
- (iv) \(\frac{x^2}{9} + \frac{y^2}{25} = 1\)

Thus, the correct answer is option (iii) \(\frac{x^2}{25} + \frac{y^2}{9} = 1\).