The equation of th directrices of the hyperbola `3x^(2)-3y^(2)-18x+12y+2=0` is

Options

A. `x=3pmsqrt((13)/(6))`
B. `x=3pmsqrt((6)/(13))`
C. `x=6pmsqrt((13)/(3))`
D. `x=6pmsqrt((3)/(13))`

Answer

To find the equation of the directrices of the hyperbola given by the equation \(3x^2 - 3y^2 - 18x + 12y + 2 = 0\), we will follow these steps:

### Step 1: Simplify the Equation
First, we simplify the equation by dividing everything by 3:

\[ x^2 - y^2 - 6x + 4y + \frac{2}{3} = 0 \]

### Step 2: Rearrange the Equation
Next, we rearrange the equation to group the \(x\) and \(y\) terms:

\[ x^2 - 6x - y^2 + 4y = -\frac{2}{3} \]

### Step 3: Complete the Square
We will complete the square for both \(x\) and \(y\):

1. For \(x^2 - 6x\):
- Take half of \(-6\), which is \(-3\), and square it to get \(9\).
- Thus, \(x^2 - 6x = (x - 3)^2 - 9\).

2. For \(-y^2 + 4y\):
- Take half of \(4\), which is \(2\), and square it to get \(4\).
- Thus, \(-y^2 + 4y = -(y^2 - 4y) = -((y - 2)^2 - 4) = -(y - 2)^2 + 4\).

Putting these back into the equation gives:

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

### Step 4: Combine and Rearrange
Now we combine the constants:

\[ (x - 3)^2 - (y - 2)^2 - 5 = -\frac{2}{3} \]

Adding \(5\) to both sides:

\[ (x - 3)^2 - (y - 2)^2 = 5 - \frac{2}{3} \]

Converting \(5\) to a fraction gives \(\frac{15}{3}\):

\[ (x - 3)^2 - (y - 2)^2 = \frac{15}{3} - \frac{2}{3} = \frac{13}{3} \]

### Step 5: Write in Standard Form
Now, we can write the equation in standard form:

\[ \frac{(x - 3)^2}{\frac{13}{3}} - \frac{(y - 2)^2}{1} = 1 \]

### Step 6: Identify Parameters
From the standard form \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), we identify:
- \(h = 3\)
- \(k = 2\)
- \(a^2 = \frac{13}{3}\) so \(a = \sqrt{\frac{13}{3}}\)
- \(b^2 = 1\) so \(b = 1\)

### Step 7: Calculate Eccentricity
The eccentricity \(e\) of the hyperbola is given by:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{\frac{13}{3}}} = \sqrt{1 + \frac{3}{13}} = \sqrt{\frac{16}{13}} = \frac{4}{\sqrt{13}} \]

### Step 8: Find the Directrices
The equations of the directrices for a hyperbola are given by:

\[ x = h \pm \frac{a}{e} \]

Calculating \(\frac{a}{e}\):

\[ \frac{a}{e} = \frac{\sqrt{\frac{13}{3}}}{\frac{4}{\sqrt{13}}} = \frac{\sqrt{13} \cdot \sqrt{13}}{4 \cdot \sqrt{3}} = \frac{13}{4\sqrt{3}} \]

Thus, the equations of the directrices are:

\[ x = 3 \pm \frac{13}{4\sqrt{3}} \]

### Final Answer
The equations of the directrices of the hyperbola are:

\[ x = 3 + \frac{13}{4\sqrt{3}} \quad \text{and} \quad x = 3 - \frac{13}{4\sqrt{3}} \]