If the eccentricity of the hyperbola is `sqrt(5)` and the distance between the foci is 12 then `b^(2)-a^(2)` is equal to (3/5) `k^(2)` where k is equal to

Options

A. 5
B. 3
C. 2
D. 6

Answer

To solve the problem step by step, we will use the properties of hyperbolas and the given information.

### Step 1: Understand the properties of the hyperbola
The standard form of a hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
The eccentricity \( e \) of a hyperbola is related to \( a \) and \( b \) by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

### Step 2: Use the given eccentricity
We are given that the eccentricity \( e = \sqrt{5} \). Therefore, we can write:
\[ \sqrt{5} = \sqrt{1 + \frac{b^2}{a^2}} \]
Squaring both sides gives:
\[ 5 = 1 + \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{b^2}{a^2} = 4 \]
Thus, we have:
\[ b^2 = 4a^2 \]

### Step 3: Use the distance between the foci
The distance between the foci of a hyperbola is given by \( 2ae \). We know the distance between the foci is 12, so:
\[ 2ae = 12 \]
Substituting \( e = \sqrt{5} \):
\[ 2a\sqrt{5} = 12 \]
Dividing both sides by 2 gives:
\[ a\sqrt{5} = 6 \]
Thus:
\[ a = \frac{6}{\sqrt{5}} = \frac{6\sqrt{5}}{5} \]

### Step 4: Find \( a^2 \)
Now we can find \( a^2 \):
\[ a^2 = \left(\frac{6\sqrt{5}}{5}\right)^2 = \frac{36 \cdot 5}{25} = \frac{180}{25} = \frac{36}{5} \]

### Step 5: Find \( b^2 \)
Using \( b^2 = 4a^2 \):
\[ b^2 = 4 \cdot \frac{36}{5} = \frac{144}{5} \]

### Step 6: Calculate \( b^2 - a^2 \)
Now we can calculate \( b^2 - a^2 \):
\[ b^2 - a^2 = \frac{144}{5} - \frac{36}{5} = \frac{144 - 36}{5} = \frac{108}{5} \]

### Step 7: Relate to the given equation
We are given that:
\[ b^2 - a^2 = \frac{3}{5} k^2 \]
Setting the two expressions equal gives:
\[ \frac{108}{5} = \frac{3}{5} k^2 \]
Multiplying both sides by 5:
\[ 108 = 3k^2 \]
Dividing by 3:
\[ k^2 = 36 \]
Taking the square root:
\[ k = 6 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{6} \]