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

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

### Step 3: Determine the orientation of the ellipse
Since \(b > a\) (5 > 4), this indicates that the major axis is along the y-axis.

### Step 4: Find the coordinates of the vertices
For an ellipse with a vertical major axis, the vertices are located at the points \((0, \pm b)\). Thus, the coordinates of the vertices are:
- \((0, 5)\)
- \((0, -5)\)

### 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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