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 : (1) `4/3` (2) `4/(sqrt(3))` (3) `2/(sqrt(3))` (4) `sqrt(3)`

Options

A. `3//4`
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: Use the length of the latus rectum
The length of the latus rectum of a hyperbola is given by the formula:
\[ L = \frac{2b^2}{a} \]
We are given that the length of the latus rectum is equal to 8. Therefore, we can set up the equation:
\[ \frac{2b^2}{a} = 8 \]
From this, we can solve for \(b^2\):
\[ 2b^2 = 8a \implies b^2 = 4a \tag{1} \]

### Step 2: Use the length of the conjugate axis
The length of the conjugate axis of a hyperbola is given by:
\[ \text{Length of conjugate axis} = 2b \]
The distance between the foci is given by:
\[ \text{Distance between foci} = 2ae \]
According to the problem, the length of the conjugate axis is equal to half the distance between the foci:
\[ 2b = \frac{1}{2} \cdot 2ae \]
This simplifies to:
\[ 2b = ae \implies b = \frac{ae}{2} \tag{2} \]

### Step 3: Substitute \(b\) from Equation (2) into Equation (1)
From Equation (2), we have \(b = \frac{ae}{2}\). Squaring both sides gives:
\[ b^2 = \left(\frac{ae}{2}\right)^2 = \frac{a^2e^2}{4} \]
Now we can substitute this expression for \(b^2\) into Equation (1):
\[ \frac{a^2e^2}{4} = 4a \]
Multiplying through by 4 to eliminate the fraction:
\[ a^2e^2 = 16a \]
Rearranging gives:
\[ a^2e^2 - 16a = 0 \]
Factoring out \(a\):
\[ a(ae^2 - 16) = 0 \]
Since \(a\) cannot be zero, we have:
\[ ae^2 - 16 = 0 \implies ae^2 = 16 \tag{3} \]

### Step 4: Use the relationship between \(e\), \(a\), and \(b\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting \(b^2 = 4a\) from Equation (1):
\[ e^2 = 1 + \frac{4a}{a^2} = 1 + \frac{4}{a} \]
Now we can substitute \(a\) from Equation (3):
From \(ae^2 = 16\), we have:
\[ a = \frac{16}{e^2} \]
Substituting this into the eccentricity equation:
\[ e^2 = 1 + \frac{4}{\frac{16}{e^2}} = 1 + \frac{4e^2}{16} = 1 + \frac{e^2}{4} \]
Multiplying through by 4 to eliminate the fraction:
\[ 4e^2 = 4 + e^2 \]
Rearranging gives:
\[ 4e^2 - e^2 = 4 \implies 3e^2 = 4 \implies e^2 = \frac{4}{3} \]
Taking the square root gives:
\[ e = \frac{2}{\sqrt{3}} \tag{4} \]

### Conclusion
Thus, the eccentricity of the hyperbola is:
\[ \boxed{\frac{2}{\sqrt{3}}} \]