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 analyze the hyperbola given by the equation \( 7y^2 - 9x^2 + 54x - 28y - 116 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 7y^2 - 9x^2 + 54x - 28y - 116 = 0 \]
We rearrange it to isolate the constant on one side:
\[ 7y^2 - 9x^2 + 54x - 28y = 116 \]

### Step 2: Grouping Terms
Next, we group the \(y\) terms and \(x\) terms:
\[ 7y^2 - 28y - 9x^2 + 54x = 116 \]

### Step 3: Completing the Square
Now, we complete the square for both \(y\) and \(x\).

**For \(y\):**
\[ 7(y^2 - 4y) \quad \text{(Take out 7)} \]
To complete the square, we take \(-4/2 = -2\) and square it to get \(4\):
\[ 7(y^2 - 4y + 4 - 4) = 7((y - 2)^2 - 4) = 7(y - 2)^2 - 28 \]

**For \(x\):**
\[ -9(x^2 - 6x) \quad \text{(Take out -9)} \]
To complete the square, we take \(6/2 = 3\) and square it to get \(9\):
\[ -9(x^2 - 6x + 9 - 9) = -9((x - 3)^2 - 9) = -9(x - 3)^2 + 81 \]

### Step 4: Substitute Back
Substituting the completed squares back into the equation gives:
\[ 7(y - 2)^2 - 28 - 9(x - 3)^2 + 81 = 116 \]
Simplifying this:
\[ 7(y - 2)^2 - 9(x - 3)^2 + 53 = 116 \]
\[ 7(y - 2)^2 - 9(x - 3)^2 = 63 \]

### Step 5: Dividing by 63
Dividing the entire equation by 63 to get it into standard form:
\[ \frac{7(y - 2)^2}{63} - \frac{9(x - 3)^2}{63} = 1 \]
This simplifies to:
\[ \frac{(y - 2)^2}{9} - \frac{(x - 3)^2}{7} = 1 \]

### Step 6: Identifying Parameters
From the standard form \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\):
- Center \((h, k) = (3, 2)\)
- \(a^2 = 9 \Rightarrow a = 3\)
- \(b^2 = 7 \Rightarrow b = \sqrt{7}\)

### Step 7: Finding Eccentricity
The eccentricity \(e\) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{7}{9}} = \sqrt{\frac{16}{9}} = \frac{4}{3} \]

### Step 8: Length of the Latus Rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} = \frac{2 \cdot 7}{3} = \frac{14}{3} \]

### Step 9: Finding the Equations of the Directrices
The equations of the directrices for a hyperbola are given by:
\[ y = k \pm \frac{a}{e} \]
Calculating:
\[ \frac{a}{e} = \frac{3}{\frac{4}{3}} = \frac{9}{4} \]
Thus, the equations of the directrices are:
\[ y = 2 + \frac{9}{4} \quad \text{and} \quad y = 2 - \frac{9}{4} \]
Calculating these gives:
\[ y = \frac{17}{4} \quad \text{and} \quad y = -\frac{1}{4} \]

### Summary of Results
1. Center: \((3, 2)\)
2. Eccentricity: \(\frac{4}{3}\)
3. Length of the Latus Rectum: \(\frac{14}{3}\)
4. Equations of the Directrices: \(y = \frac{17}{4}\) and \(y = -\frac{1}{4}\)