Find the centre, foci and the eccentricity of the hyperbola. `11x^(2) -25 y^(2) -44x + 50 y-256 = 0`

Answer

To find the center, foci, and eccentricity of the hyperbola given by the equation \(11x^2 - 25y^2 - 44x + 50y - 256 = 0\), we will follow these steps:

### Step 1: Rearrange the equation
First, we will rearrange the equation to group the \(x\) and \(y\) terms together.

\[ 11x^2 - 44x - 25y^2 + 50y = 256 \]

### Step 2: Complete the square for \(x\) terms
Next, we will complete the square for the \(x\) terms. We have:

\[ 11(x^2 - 4x) - 25(y^2 - 2y) = 256 \]

To complete the square for \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

So we can rewrite it as:
\[ 11((x - 2)^2 - 4) = 11(x - 2)^2 - 44 \]

### Step 3: Complete the square for \(y\) terms
Now, we will complete the square for the \(y\) terms:
\[ -25(y^2 - 2y) = -25((y - 1)^2 - 1) = -25(y - 1)^2 + 25 \]

### Step 4: Substitute back into the equation
Substituting back into the equation gives us:
\[ 11((x - 2)^2 - 4) - 25((y - 1)^2 - 1) = 256 \]
\[ 11(x - 2)^2 - 44 - 25(y - 1)^2 + 25 = 256 \]
\[ 11(x - 2)^2 - 25(y - 1)^2 - 19 = 256 \]
\[ 11(x - 2)^2 - 25(y - 1)^2 = 275 \]

### Step 5: Divide by 275
Now, we divide the entire equation by 275 to get it in standard form:
\[ \frac{11(x - 2)^2}{275} - \frac{25(y - 1)^2}{275} = 1 \]
This simplifies to:
\[ \frac{(x - 2)^2}{25} - \frac{(y - 1)^2}{11} = 1 \]

### Step 6: Identify the center, \(a^2\), and \(b^2\)
From the standard form of the hyperbola:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
we can identify:
- Center \((h, k) = (2, 1)\)
- \(a^2 = 25 \Rightarrow a = 5\)
- \(b^2 = 11 \Rightarrow b = \sqrt{11}\)

### Step 7: Calculate the foci
The foci of the hyperbola are given by:
\[ c = \sqrt{a^2 + b^2} = \sqrt{25 + 11} = \sqrt{36} = 6 \]
Thus, the foci are located at:
\[ (h \pm c, k) = (2 \pm 6, 1) = (8, 1) \text{ and } (-4, 1) \]

### Step 8: Calculate the eccentricity
The eccentricity \(e\) of the hyperbola is given by:
\[ e = \frac{c}{a} = \frac{6}{5} \]

### Final Results
- **Center**: \((2, 1)\)
- **Foci**: \((8, 1)\) and \((-4, 1)\)
- **Eccentricity**: \(\frac{6}{5}\)