If the eccentricity of the hyperbola `(x^(2))/(16)-(y^(2))/(b^(2))=-1` is `(5)/(4)`, then `b^(2)` is equal to

Options

A. `(256)/(9)`
B. `(16)/(3)`
C. 9
D. 3

Answer

To solve the problem, we need to find the value of \( b^2 \) for the hyperbola given by the equation:

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

We know that the eccentricity \( e \) of a hyperbola is given by the formula:

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

where \( a^2 \) and \( b^2 \) are the squares of the semi-major and semi-minor axes, respectively. In our case, \( a^2 = 16 \) (since the hyperbola is in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)).

### Step 1: Identify the eccentricity
We are given that the eccentricity \( e \) is \( \frac{5}{4} \).

### Step 2: Set up the equation for eccentricity
Using the formula for eccentricity, we can substitute the known values:

\[ \frac{5}{4} = \sqrt{1 + \frac{b^2}{16}} \]

### Step 3: Square both sides
To eliminate the square root, we square both sides:

\[ \left(\frac{5}{4}\right)^2 = 1 + \frac{b^2}{16} \]

Calculating the left side:

\[ \frac{25}{16} = 1 + \frac{b^2}{16} \]

### Step 4: Isolate \( \frac{b^2}{16} \)
Now, we can isolate \( \frac{b^2}{16} \):

\[ \frac{25}{16} - 1 = \frac{b^2}{16} \]

Convert 1 to a fraction with a denominator of 16:

\[ \frac{25}{16} - \frac{16}{16} = \frac{b^2}{16} \]

This simplifies to:

\[ \frac{9}{16} = \frac{b^2}{16} \]

### Step 5: Solve for \( b^2 \)
Now, we can multiply both sides by 16 to solve for \( b^2 \):

\[ b^2 = 9 \]

### Final Answer
Thus, the value of \( b^2 \) is:

\[ \boxed{9} \]