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

Options

A. `(pm5,0)`
B. `(pm4,0)`
C. `(0,pm5)`
D. `(0,pm4)`

Answer

To find the vertices 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 \(a^2\) and \(b^2\)
The equation of the ellipse is in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). From the given equation, we can identify:
- \(a^2 = 16\)
- \(b^2 = 25\)

### Step 2: Calculate \(a\) and \(b\)
Now, we take the square root of \(a^2\) and \(b^2\) to find \(a\) and \(b\):
- \(a = \sqrt{16} = 4\)
- \(b = \sqrt{25} = 5\)

### Step 3: Determine the orientation of the ellipse
Since \(b > a\) (5 > 4), this indicates that the ellipse is vertically oriented. For a vertically oriented ellipse, the vertices are located at the points \((0, \pm b)\).

### Step 4: Find the coordinates of the vertices
Using the value of \(b\):
- The vertices are at \((0, +b)\) and \((0, -b)\).
- Therefore, the vertices are \((0, 5)\) and \((0, -5)\).

### Final Answer
The vertices of the ellipse \(\frac{x^2}{16} + \frac{y^2}{25} = 1\) are \((0, 5)\) and \((0, -5)\).

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