The eccentricity of the ellipse `(x^2)/25+(y^2)/9=1` is ,

Options

A. a)`4/5`
B. `-4/5`
C. `3/5`
D. d)`-3/5`

Answer

To find the eccentricity of the ellipse given by the equation \(\frac{x^2}{25} + \frac{y^2}{9} = 1\), we can follow these steps:

### Step 1: Identify the values of \(a\) and \(b\).
The standard form of the ellipse is given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). From the given equation, we can identify:
- \(a^2 = 25\) → \(a = \sqrt{25} = 5\)
- \(b^2 = 9\) → \(b = \sqrt{9} = 3\)

### Step 2: Use the formula for eccentricity.
The eccentricity \(e\) of an ellipse is calculated using the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

### Step 3: Substitute the values of \(a\) and \(b\) into the formula.
Now, substituting the values we found:
\[ e = \sqrt{1 - \frac{3^2}{5^2}} = \sqrt{1 - \frac{9}{25}} \]

### Step 4: Simplify the expression inside the square root.
To simplify:
\[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} \]

### Step 5: Calculate the square root.
Taking the square root gives:
\[ e = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \]

### Conclusion:
Thus, the eccentricity of the ellipse is \(\frac{4}{5}\).