If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is

Options

A. 2
B. `(13)/(6)`
C. `(13)/(8)`
D. `(13)/(12)`

Answer

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

### Step 1: Understand the given information
We are given:
- Length of the conjugate axis = 5
- Distance between the foci = 13

### Step 2: Relate the conjugate axis to 'b'
The length of the conjugate axis of a hyperbola is given by \(2b\). Therefore, we can write:
\[ 2b = 5 \implies b = \frac{5}{2} \]

### Step 3: Relate the distance between the foci to 'a' and 'e'
The distance between the foci of a hyperbola is given by \(2ae\). Thus, we have:
\[ 2ae = 13 \implies ae = \frac{13}{2} \]

### Step 4: Use the relationship between \(a\), \(b\), and \(e\)
For a hyperbola, the relationship between \(a\), \(b\), and \(e\) is given by:
\[ b^2 = a^2(e^2 - 1) \]

### Step 5: Substitute the values of \(b\) and \(ae\)
First, we calculate \(b^2\):
\[ b^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]

Now, substituting \(ae = \frac{13}{2}\) into the equation \(e = \frac{ae}{a}\):
\[ e = \frac{\frac{13}{2}}{a} \]

### Step 6: Substitute \(e\) into the relationship
Substituting \(e\) into the equation \(b^2 = a^2(e^2 - 1)\):
\[ \frac{25}{4} = a^2\left(\left(\frac{\frac{13}{2}}{a}\right)^2 - 1\right) \]

### Step 7: Simplify the equation
Calculating \(e^2\):
\[ e^2 = \left(\frac{\frac{13}{2}}{a}\right)^2 = \frac{169}{4a^2} \]
Thus, we have:
\[ \frac{25}{4} = a^2\left(\frac{169}{4a^2} - 1\right) \]
This simplifies to:
\[ \frac{25}{4} = \frac{169 - 4a^2}{4} \]

### Step 8: Clear the fraction
Multiplying through by 4 gives:
\[ 25 = 169 - 4a^2 \]

### Step 9: Solve for \(a^2\)
Rearranging gives:
\[ 4a^2 = 169 - 25 = 144 \implies a^2 = \frac{144}{4} = 36 \]

### Step 10: Find \(a\)
Taking the square root:
\[ a = 6 \]

### Step 11: Find \(e\)
Now substituting \(a\) back to find \(e\):
\[ e = \frac{ae}{a} = \frac{\frac{13}{2}}{6} = \frac{13}{12} \]

### Final Answer
The eccentricity \(e\) of the hyperbola is:
\[ \boxed{\frac{13}{12}} \]