The length of latus-rectum of the hyperbola `x^(2) - 2y ^(2) - 2x + 8y - 1 = 0 ` is
(i) `sqrt(6)`
(ii) `4 sqrt(3)`
(iii) 4
(iv) 3

Options

A. `sqrt(6)`
B. `4 sqrt(3)`
C. 4
D. 3

Answer

To find the length of the latus rectum of the hyperbola given by the equation \(x^2 - 2y^2 - 2x + 8y - 1 = 0\), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ x^2 - 2y^2 - 2x + 8y - 1 = 0 \]
Rearranging gives:
\[ x^2 - 2x - 2y^2 + 8y - 1 = 0 \]

### Step 2: Complete the square for \(x\) and \(y\)
For \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

For \(y\):
\[ -2y^2 + 8y = -2(y^2 - 4y) = -2((y - 2)^2 - 4) = -2(y - 2)^2 + 8 \]

Substituting these back into the equation gives:
\[ (x - 1)^2 - 1 - 2(y - 2)^2 + 8 - 1 = 0 \]
This simplifies to:
\[ (x - 1)^2 - 2(y - 2)^2 + 6 = 0 \]

### Step 3: Isolate the terms
Rearranging gives:
\[ (x - 1)^2 - 2(y - 2)^2 = -6 \]
Dividing through by \(-6\) to get it into standard form:
\[ -\frac{(x - 1)^2}{6} + \frac{(y - 2)^2}{3} = 1 \]
This can be rewritten as:
\[ \frac{(y - 2)^2}{3} - \frac{(x - 1)^2}{6} = 1 \]

### Step 4: Identify \(a^2\) and \(b^2\)
From the standard form of the hyperbola \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\), we identify:
- \(a^2 = 3\)
- \(b^2 = 6\)

### Step 5: Calculate the length of the latus rectum
The formula for the length of the latus rectum \(L\) of a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values we found:
\[ L = \frac{2 \cdot 6}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

### Conclusion
Thus, the length of the latus rectum of the hyperbola is:
\[ \boxed{4\sqrt{3}} \]