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

Options

A. `(4)/(3)`
B. `(4)/(sqrt(3))`
C. `(2)/(sqrt(3))`
D. `2 sqrt(3)`

Answer

To find the eccentricity of the hyperbola given the latus rectum and the relationship between the conjugate axis and the distance between the foci, we can follow these steps:

### Step 1: Understand the given information
- The length of the latus rectum (L) is given as 8.
- The length of the conjugate axis (2b) is equal to half the distance between the foci (2c), which means:
\[ 2b = \frac{1}{2} \times 2c \implies 2b = c \]

### Step 2: Use the formula for the latus rectum
For a hyperbola, the length of the latus rectum is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the value of L:
\[ 8 = \frac{2b^2}{a} \]
From this, we can derive:
\[ b^2 = 4a \]

### Step 3: Relate b to c
From the relationship derived earlier, we have:
\[ c = 2b \]
We also know that for hyperbolas, the relationship between a, b, and c is given by:
\[ c^2 = a^2 + b^2 \]
Substituting \(c = 2b\) into this equation:
\[ (2b)^2 = a^2 + b^2 \implies 4b^2 = a^2 + b^2 \]
This simplifies to:
\[ 4b^2 - b^2 = a^2 \implies 3b^2 = a^2 \implies b^2 = \frac{a^2}{3} \]

### Step 4: Substitute b² into the equation from latus rectum
Now we have two expressions for \(b^2\):
1. From the latus rectum: \(b^2 = 4a\)
2. From the relationship with a: \(b^2 = \frac{a^2}{3}\)

Setting these equal to each other:
\[ 4a = \frac{a^2}{3} \]
Cross-multiplying gives:
\[ 12a = a^2 \implies a^2 - 12a = 0 \]
Factoring out \(a\):
\[ a(a - 12) = 0 \]
Thus, \(a = 0\) or \(a = 12\). Since \(a\) cannot be 0, we have:
\[ a = 12 \]

### Step 5: Find b²
Substituting \(a = 12\) back into the equation \(b^2 = 4a\):
\[ b^2 = 4 \times 12 = 48 \]

### Step 6: Find c²
Now, using the relationship \(c^2 = a^2 + b^2\):
\[ c^2 = 12^2 + 48 = 144 + 48 = 192 \]

### Step 7: Find eccentricity
The eccentricity \(e\) is given by:
\[ e = \frac{c}{a} \]
Calculating \(c\):
\[ c = \sqrt{192} = 8\sqrt{3} \]
Thus,
\[ e = \frac{8\sqrt{3}}{12} = \frac{2\sqrt{3}}{3} \]

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