If equation of hyperbola is `4(2y -x -3)^(2) -9(2x + y - 1)^(2) = 80`, then

Options

A. Eccentricity is `(sqrt(13))/(3)`
B. Centre of hyperbola is `(-(7)/(5),-(7)/(5))`
C. Transverse axis is `2x + y -1 = 0`
D. Conjugate axis is `x -2y + 3 = 0`

Answer

To solve the given hyperbola equation \(4(2y - x - 3)^{2} - 9(2x + y - 1)^{2} = 80\), we will follow these steps:

### Step 1: Simplify the equation
First, we will divide the entire equation by 80 to simplify it.

\[ \frac{4(2y - x - 3)^{2}}{80} - \frac{9(2x + y - 1)^{2}}{80} = 1 \]

This simplifies to:

\[ \frac{(2y - x - 3)^{2}}{20} - \frac{(2x + y - 1)^{2}}{\frac{80}{9}} = 1 \]

### Step 2: Identify the standard form
The standard form of a hyperbola is given by:

\[ \frac{(y - k)^{2}}{a^{2}} - \frac{(x - h)^{2}}{b^{2}} = 1 \]

From our equation, we can identify:

- \(a^2 = 20\)
- \(b^2 = \frac{80}{9}\)

### Step 3: Find the center of the hyperbola
To find the center of the hyperbola, we need to determine the points where the expressions \(2y - x - 3 = 0\) and \(2x + y - 1 = 0\) intersect.

1. From \(2y - x - 3 = 0\), we can express \(y\) in terms of \(x\):

\[ 2y = x + 3 \implies y = \frac{x + 3}{2} \]

2. Substitute \(y\) into the second equation \(2x + y - 1 = 0\):

\[ 2x + \frac{x + 3}{2} - 1 = 0 \]

Multiply through by 2 to eliminate the fraction:

\[ 4x + x + 3 - 2 = 0 \implies 5x + 1 = 0 \implies x = -\frac{1}{5} \]

3. Substitute \(x = -\frac{1}{5}\) back into \(y = \frac{x + 3}{2}\):

\[ y = \frac{-\frac{1}{5} + 3}{2} = \frac{\frac{14}{5}}{2} = \frac{7}{5} \]

Thus, the center of the hyperbola is at the point \(\left(-\frac{1}{5}, \frac{7}{5}\right)\).

### Step 4: Determine the transverse and conjugate axes
The transverse axis is along the direction of the \(y\)-axis (since the \(y\) term is positive in the standard form), and the conjugate axis is along the \(x\)-axis.

### Step 5: Calculate the eccentricity
The eccentricity \(e\) of a hyperbola is given by:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Substituting the values of \(a^2\) and \(b^2\):

\[ e = \sqrt{1 + \frac{\frac{80}{9}}{20}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Final Result
The final answers are:
- Center: \(\left(-\frac{1}{5}, \frac{7}{5}\right)\)
- Eccentricity: \(e = \frac{\sqrt{13}}{3}\)