The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

Options

A. `4/3`
B. `4/sqrt3`
C. `2/sqrt3`
D. none of these

Explanation

To find the eccentricity of the hyperbola given the conditions in the problem, we can follow these steps:

### Step 1: Understand the given information
We are given:
- Length of the latus rectum (L) = 8
- Length of the conjugate axis (2b) = half of the distance between the foci (2ae)

### Step 2: Use the formula for the length of the latus rectum
The formula for the length of the latus rectum of a hyperbola is:
\[ L = \frac{2b^2}{a} \]
Given that \(L = 8\), we can write:
\[ \frac{2b^2}{a} = 8 \]
From this, we can derive:
\[ b^2 = 4a \]

### Step 3: Relate the conjugate axis and the distance between the foci
The length of the conjugate axis is given by \(2b\), and the distance between the foci is \(2ae\). According to the problem:
\[ 2b = \frac{1}{2}(2ae) \implies 2b = ae \]
This simplifies to:
\[ b = \frac{ae}{2} \]

### Step 4: Substitute the expression for b into the equation from Step 2
We already have \(b^2 = 4a\). Now, substituting \(b = \frac{ae}{2}\) into this equation:
\[ \left(\frac{ae}{2}\right)^2 = 4a \]
This expands to:
\[ \frac{a^2e^2}{4} = 4a \]
Multiplying both sides by 4 gives:
\[ a^2e^2 = 16a \]

### Step 5: Rearranging the equation
We can rearrange this to:
\[ a^2e^2 - 16a = 0 \]
Factoring out \(a\):
\[ a(a e^2 - 16) = 0 \]
Since \(a \neq 0\), we have:
\[ ae^2 - 16 = 0 \implies ae^2 = 16 \implies a = \frac{16}{e^2} \]

### Step 6: Substitute \(a\) back to find \(b\)
Now we can substitute \(a = \frac{16}{e^2}\) back into the equation for \(b\):
\[ b^2 = 4a = 4 \left(\frac{16}{e^2}\right) = \frac{64}{e^2} \]

### Step 7: Use the relationship between \(a\), \(b\), and \(e\)
We know that:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting the values of \(b^2\) and \(a\):
\[ e^2 = 1 + \frac{\frac{64}{e^2}}{\left(\frac{16}{e^2}\right)^2} \]
This simplifies to:
\[ e^2 = 1 + \frac{64}{\frac{256}{e^4}} = 1 + \frac{64e^4}{256} = 1 + \frac{e^4}{4} \]

### Step 8: Rearranging the equation
Multiplying through by 4 to eliminate the fraction:
\[ 4e^2 = 4 + e^4 \]
Rearranging gives:
\[ e^4 - 4e^2 + 4 = 0 \]

### Step 9: Let \(x = e^2\)
Letting \(x = e^2\), we have:
\[ x^2 - 4x + 4 = 0 \]
Factoring gives:
\[ (x - 2)^2 = 0 \implies x = 2 \]
Thus,
\[ e^2 = 2 \implies e = \sqrt{2} \]

### Final Answer
The eccentricity \(e\) of the hyperbola is:
\[ \boxed{\sqrt{2}} \]