Which of the following is /are true about the hyperbola `7y^(2) - 9x^(2) + 54x - 28y - 116 = 0 `

Options

A. coordinates of the centre are `(3, 2)`
B. eccentricity is `4/sqrt7`
C. length of latus rectum is `14//3`
D. equation of directrices are `y = 2 pm (3 sqrt7)/4`

Answer

To determine the properties of the hyperbola given by the equation \( 7y^2 - 9x^2 + 54x - 28y - 116 = 0 \), we will first rewrite the equation in standard form.

### Step 1: Rearrange the equation
We start with the equation:
\[ 7y^2 - 9x^2 + 54x - 28y - 116 = 0 \]
We can rearrange it as:
\[ 7y^2 - 28y - 9x^2 + 54x - 116 = 0 \]

### Step 2: Group the terms
Next, we group the \(y\) terms and the \(x\) terms:
\[ 7(y^2 - 4y) - 9(x^2 - 6x) - 116 = 0 \]

### Step 3: Complete the square for \(y\)
To complete the square for the \(y\) terms:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Substituting this back into the equation gives:
\[ 7((y - 2)^2 - 4) - 9(x^2 - 6x) - 116 = 0 \]
This simplifies to:
\[ 7(y - 2)^2 - 28 - 9(x^2 - 6x) - 116 = 0 \]

### Step 4: Complete the square for \(x\)
Now, we complete the square for the \(x\) terms:
\[ x^2 - 6x = (x - 3)^2 - 9 \]
Substituting this back gives:
\[ 7(y - 2)^2 - 28 - 9((x - 3)^2 - 9) - 116 = 0 \]
This simplifies to:
\[ 7(y - 2)^2 - 28 - 9(x - 3)^2 + 81 - 116 = 0 \]
Combining constants:
\[ 7(y - 2)^2 - 9(x - 3)^2 - 63 = 0 \]

### Step 5: Move constant to the other side
Now, we can move the constant to the other side:
\[ 7(y - 2)^2 - 9(x - 3)^2 = 63 \]

### Step 6: Divide by 63
Dividing the entire equation by 63 gives:
\[ \frac{(y - 2)^2}{9} - \frac{(x - 3)^2}{7} = 1 \]

### Step 7: Identify the standard form
Now we have the equation in the standard form of a hyperbola:
\[ \frac{(y - 2)^2}{9} - \frac{(x - 3)^2}{7} = 1 \]
This indicates that the hyperbola opens upwards and downwards.

### Conclusion
From the standard form, we can conclude:
- The center of the hyperbola is at \((3, 2)\).
- The semi-major axis length is \(3\) (since \(\sqrt{9} = 3\)).
- The semi-minor axis length is \(\sqrt{7}\).
- The hyperbola opens vertically.