Eccentricity of the conic ` 7y^(2) - 9x^(2) +54 x - 28y -116 =0 ` is

Options

A. `(3)/(2)`
B. ` ( 4)/(3)`
C. ` (3)/(4)`
D. None of these

Answer

To find the eccentricity of the conic 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:
\[ 7y^2 - 9x^2 + 54x - 28y - 116 = 0 \]
Rearranging it gives:
\[ 7y^2 - 28y - 9x^2 + 54x = 116 \]

### Step 2: Completing the square for \(y\)
We will complete the square for the \(y\) terms:
\[ 7(y^2 - 4y) = 7[(y - 2)^2 - 4] = 7(y - 2)^2 - 28 \]
So, substituting back, we have:
\[ 7(y - 2)^2 - 28 - 9x^2 + 54x = 116 \]
This simplifies to:
\[ 7(y - 2)^2 - 9x^2 + 54x - 28 = 116 \]

### Step 3: Completing the square for \(x\)
Next, we complete the square for the \(x\) terms:
\[ -9(x^2 - 6x) = -9[(x - 3)^2 - 9] = -9(x - 3)^2 + 81 \]
Substituting this back gives:
\[ 7(y - 2)^2 - 9(x - 3)^2 + 81 - 28 = 116 \]
This simplifies to:
\[ 7(y - 2)^2 - 9(x - 3)^2 + 53 = 116 \]
So we have:
\[ 7(y - 2)^2 - 9(x - 3)^2 = 63 \]

### Step 4: Dividing by 63
Now we divide the entire equation by 63:
\[ \frac{7(y - 2)^2}{63} - \frac{9(x - 3)^2}{63} = 1 \]
This simplifies to:
\[ \frac{(y - 2)^2}{9} - \frac{(x - 3)^2}{7} = 1 \]

### Step 5: Identifying the conic type
This is the standard form of a hyperbola:
\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]
where \(a^2 = 9\) and \(b^2 = 7\).

### Step 6: Calculating the eccentricity
For a hyperbola, the eccentricity \(e\) is given by:
\[ e = \sqrt{1 + \frac{a^2}{b^2}} \]
Substituting \(a^2 = 9\) and \(b^2 = 7\):
\[ e = \sqrt{1 + \frac{9}{7}} = \sqrt{\frac{7 + 9}{7}} = \sqrt{\frac{16}{7}} = \frac{4}{\sqrt{7}} \]

### Final Answer
Thus, the eccentricity of the conic is:
\[ e = \frac{4}{\sqrt{7}} \]