Find the centre, eccentricity and length of axes of the hyperbola `3x^(2)-5y^(2)-6x+20y-32=0`.

Answer

To solve the hyperbola equation \(3x^2 - 5y^2 - 6x + 20y - 32 = 0\) and find the center, eccentricity, and lengths of the axes, we will follow these steps:

### Step 1: Rearranging the equation
We start with the given equation:
\[ 3x^2 - 5y^2 - 6x + 20y - 32 = 0 \]
Rearranging gives:
\[ 3x^2 - 6x - 5y^2 + 20y = 32 \]

### Step 2: Completing the square for \(x\) and \(y\)
Next, we complete the square for the \(x\) and \(y\) terms.

**For \(x\):**
\[ 3(x^2 - 2x) = 3((x - 1)^2 - 1) = 3(x - 1)^2 - 3 \]

**For \(y\):**
\[ -5(y^2 - 4y) = -5((y - 2)^2 - 4) = -5(y - 2)^2 + 20 \]

Substituting these back into the equation gives:
\[ 3((x - 1)^2 - 1) - 5((y - 2)^2 - 4) = 32 \]
This simplifies to:
\[ 3(x - 1)^2 - 5(y - 2)^2 - 3 + 20 = 32 \]
\[ 3(x - 1)^2 - 5(y - 2)^2 + 17 = 32 \]
\[ 3(x - 1)^2 - 5(y - 2)^2 = 15 \]

### Step 3: Dividing by 15 to get the standard form
Now, we divide the entire equation by 15:
\[ \frac{3(x - 1)^2}{15} - \frac{5(y - 2)^2}{15} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{5} - \frac{(y - 2)^2}{3} = 1 \]

### Step 4: Identifying the center, \(a^2\), and \(b^2\)
From the standard form \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), we identify:
- Center \((h, k) = (1, 2)\)
- \(a^2 = 5\) and \(b^2 = 3\)

### Step 5: Finding the eccentricity
The eccentricity \(e\) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{5}} = \sqrt{\frac{8}{5}} = \frac{2\sqrt{2}}{\sqrt{5}} \]

### Step 6: Finding the lengths of the axes
- Length of the transverse axis \(= 2a = 2\sqrt{5}\)
- Length of the conjugate axis \(= 2b = 2\sqrt{3}\)

### Final Results
- **Center:** \((1, 2)\)
- **Eccentricity:** \(\sqrt{\frac{8}{5}} = \frac{2\sqrt{2}}{\sqrt{5}}\)
- **Length of Transverse Axis:** \(2\sqrt{5}\)
- **Length of Conjugate Axis:** \(2\sqrt{3}\)