If `3x^(2) - 5y^(2) - 6x + 20 y - 32 = 0` represents a hyperbola, then the co-ordinates of foci are

Options

A. `(pm 2sqrt(2), 0)`
B. `(1 pm 2sqrt(2), 2)`
C. `(0, pm 2sqrt(2))`
D. `(1, 2)`

Answer

To find the coordinates of the foci of the hyperbola represented by the equation \(3x^2 - 5y^2 - 6x + 20y - 32 = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation to group the \(x\) and \(y\) terms together:
\[ 3x^2 - 6x - 5y^2 + 20y = 32 \]

### Step 2: Completing the Square for \(x\)
For the \(x\) terms, factor out the coefficient of \(x^2\) (which is 3):
\[ 3(x^2 - 2x) - 5y^2 + 20y = 32 \]
Now, complete the square for \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Substituting back, we have:
\[ 3((x - 1)^2 - 1) - 5y^2 + 20y = 32 \]
This simplifies to:
\[ 3(x - 1)^2 - 3 - 5y^2 + 20y = 32 \]

### Step 3: Completing the Square for \(y\)
Now, for the \(y\) terms, factor out \(-5\):
\[ -5(y^2 - 4y) = -5((y - 2)^2 - 4) = -5(y - 2)^2 + 20 \]
Substituting this back, we have:
\[ 3(x - 1)^2 - 3 - 5(y - 2)^2 + 20 = 32 \]
This simplifies to:
\[ 3(x - 1)^2 - 5(y - 2)^2 + 17 = 32 \]

### Step 4: Final Form of the Hyperbola
Now, isolate the hyperbola equation:
\[ 3(x - 1)^2 - 5(y - 2)^2 = 32 - 17 \]
\[ 3(x - 1)^2 - 5(y - 2)^2 = 15 \]
Dividing through by 15 gives:
\[ \frac{(x - 1)^2}{5} - \frac{(y - 2)^2}{3} = 1 \]

### Step 5: Identifying Parameters
From the standard form of the hyperbola \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), we identify:
- \(h = 1\)
- \(k = 2\)
- \(a^2 = 5\) ⇒ \(a = \sqrt{5}\)
- \(b^2 = 3\) ⇒ \(b = \sqrt{3}\)

### Step 6: Finding the Eccentricity
The eccentricity \(e\) of the 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 7: Finding the Foci
The coordinates of the foci are given by:
\[ (h \pm ae, k) = \left(1 \pm \sqrt{5} \cdot \frac{2\sqrt{2}}{\sqrt{5}}, 2\right) \]
This simplifies to:
\[ (1 \pm 2\sqrt{2}, 2) \]

### Final Answer
Thus, the coordinates of the foci are:
\[ (1 + 2\sqrt{2}, 2) \quad \text{and} \quad (1 - 2\sqrt{2}, 2) \]

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