The centre of the hyperbola `x^(2)-y^(2)-4x-2y-8=0` is

Options

A. (2,2)
B. (2,-2)
C. (2,1)
D. (2,-1)

Answer

To find the center of the hyperbola given by the equation \( x^2 - y^2 - 4x - 2y - 8 = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
Start with the given equation:
\[ x^2 - y^2 - 4x - 2y - 8 = 0 \]
Rearranging gives:
\[ x^2 - 4x - y^2 - 2y = 8 \]

### Step 2: Complete the square for \(x\)
For the \(x\) terms \(x^2 - 4x\), we complete the square:
- Take half of the coefficient of \(x\) (which is \(-4\)), square it, and add/subtract it:
\[ x^2 - 4x = (x - 2)^2 - 4 \]
Now substitute this back into the equation:
\[ (x - 2)^2 - 4 - y^2 - 2y = 8 \]
This simplifies to:
\[ (x - 2)^2 - y^2 - 2y - 4 = 8 \]

### Step 3: Complete the square for \(y\)
Now, for the \(y\) terms \(-y^2 - 2y\), we can factor out \(-1\) and complete the square:
\[ -y^2 - 2y = - (y^2 + 2y) = -((y + 1)^2 - 1) = -(y + 1)^2 + 1 \]
Substituting this back gives:
\[ (x - 2)^2 - (-(y + 1)^2 + 1) - 4 = 8 \]
This simplifies to:
\[ (x - 2)^2 - (y + 1)^2 + 1 - 4 = 8 \]
or:
\[ (x - 2)^2 - (y + 1)^2 - 3 = 8 \]

### Step 4: Simplify the equation
Now, we can simplify the equation:
\[ (x - 2)^2 - (y + 1)^2 = 11 \]

### Step 5: Write in standard form
Now we can write the equation in standard form:
\[ \frac{(x - 2)^2}{11} - \frac{(y + 1)^2}{11} = 1 \]

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

Thus, the center of the hyperbola is:
\[ \boxed{(2, -1)} \]

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