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

Options

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

Answer

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

### Step 1: Rewrite the equation in standard form
The standard form of a hyperbola is given by:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

We start with the equation:

\[ 16x^2 - 9y^2 = 1 \]

To rewrite it in the standard form, we divide the entire equation by 1:

\[ \frac{16x^2}{1} - \frac{9y^2}{1} = 1 \]

Now, we can express it as:

\[ \frac{x^2}{\frac{1}{16}} - \frac{y^2}{\frac{1}{9}} = 1 \]

### Step 2: Identify \( a^2 \) and \( b^2 \)
From the rewritten equation, we can identify:

\[ a^2 = \frac{1}{16} \quad \text{and} \quad b^2 = \frac{1}{9} \]

### Step 3: Calculate \( a \) and \( b \)
Now, we can find \( a \) and \( b \):

\[ a = \sqrt{\frac{1}{16}} = \frac{1}{4} \quad \text{and} \quad b = \sqrt{\frac{1}{9}} = \frac{1}{3} \]

### Step 4: Use the formula for eccentricity
The formula for the eccentricity \( e \) of a hyperbola is given by:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Substituting the values of \( a^2 \) and \( b^2 \):

\[ e = \sqrt{1 + \frac{\frac{1}{9}}{\frac{1}{16}}} \]

### Step 5: Simplify the expression
Calculating \( \frac{b^2}{a^2} \):

\[ \frac{\frac{1}{9}}{\frac{1}{16}} = \frac{1}{9} \times \frac{16}{1} = \frac{16}{9} \]

Now substituting this back into the formula for \( e \):

\[ e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{9 + 16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \]

### Final Answer
Thus, the eccentricity of the hyperbola \( 16x^2 - 9y^2 = 1 \) is:

\[ \boxed{\frac{5}{3}} \]