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. (a) `4//sqrt3`
B. (b) `2//sqrt3`
C. (c) `sqrt3`
D. (d) `4//3`

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 know:
- The length of the latus rectum \( L = 8 \)
- The length of the conjugate axis \( 2b \) is equal to half the distance between the foci.

### Step 2: Relate the latus rectum to \( a \) and \( b \)
The formula for the length of the latus rectum of a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the given value of \( L \):
\[ \frac{2b^2}{a} = 8 \]
This simplifies to:
\[ b^2 = 4a \]

### Step 3: Relate the conjugate axis to the distance between the foci
The distance between the foci of a hyperbola is given by \( 2ae \), where \( e \) is the eccentricity. The problem states that:
\[ 2b = \frac{1}{2}(2ae) \]
This simplifies to:
\[ 2b = ae \]
Thus:
\[ b = \frac{ae}{2} \]

### Step 4: Substitute \( b \) in terms of \( a \) and \( e \)
Now we can substitute \( b \) from the previous equation into the equation \( b^2 = 4a \):
\[ \left(\frac{ae}{2}\right)^2 = 4a \]
This expands to:
\[ \frac{a^2e^2}{4} = 4a \]
Multiplying both sides by 4 to eliminate the fraction:
\[ a^2e^2 = 16a \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ a^2e^2 - 16a = 0 \]
Factoring out \( a \):
\[ a(a e^2 - 16) = 0 \]
Since \( a \neq 0 \), we have:
\[ ae^2 - 16 = 0 \]
Thus:
\[ ae^2 = 16 \]

### Step 6: Substitute \( b^2 \) into the eccentricity formula
We already have \( b^2 = 4a \). Now we can express \( e^2 \) in terms of \( a \):
Using the relationship for eccentricity:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting \( b^2 = 4a \):
\[ e^2 = 1 + \frac{4a}{a^2} = 1 + \frac{4}{a} \]

### Step 7: Substitute \( ae^2 = 16 \) into the eccentricity formula
From \( ae^2 = 16 \):
\[ e^2 = \frac{16}{a} \]
Now we can equate the two expressions for \( e^2 \):
\[ \frac{16}{a} = 1 + \frac{4}{a} \]
Multiplying through by \( a \) gives:
\[ 16 = a + 4 \]
Thus:
\[ a = 12 \]

### Step 8: Calculate \( e^2 \)
Substituting \( a = 12 \) back into \( ae^2 = 16 \):
\[ 12e^2 = 16 \implies e^2 = \frac{16}{12} = \frac{4}{3} \]

### Step 9: Find \( e \)
Taking the square root gives:
\[ e = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Final Answer
The eccentricity \( e \) of the hyperbola is:
\[ \frac{2}{\sqrt{3}} \]