The eccentricity of `x^(2)/9-y^(2)/(16)=1` is

Options

A. `(17)/(16)`
B. `5/4`
C. `5/3`
D. `sqrt7/4`

Answer

To find the eccentricity of the hyperbola given by the equation \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \), we can follow these steps:

### Step 1: Identify the standard form of the hyperbola
The given equation is \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \). This is in the standard form of a hyperbola, which is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

### Step 2: Determine the values of \( a^2 \) and \( b^2 \)
From the equation \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \), we can identify:
- \( a^2 = 9 \)
- \( b^2 = 16 \)

### Step 3: Calculate \( a \) and \( b \)
Now, we find \( a \) and \( b \):
- \( a = \sqrt{9} = 3 \)
- \( b = \sqrt{16} = 4 \)

### Step 4: Use the formula for eccentricity
The formula for the eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

### Step 5: Substitute the values of \( a^2 \) and \( b^2 \)
Substituting the values we found:
\[ e = \sqrt{1 + \frac{16}{9}} \]

### Step 6: Simplify the expression
First, calculate \( \frac{16}{9} \):
\[ e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{9}{9} + \frac{16}{9}} = \sqrt{\frac{25}{9}} \]

### Step 7: Final calculation of eccentricity
Now, we can simplify:
\[ e = \frac{\sqrt{25}}{\sqrt{9}} = \frac{5}{3} \]

### Conclusion
Thus, the eccentricity of the hyperbola \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \) is \( \frac{5}{3} \).