The equation of directrices of the hyperbola ` 5x^(2) -4y^(2) -30x -8y -39 =0` are

Options

A. `x= pm 9//5`
B. `x =3 pm 8//3`
C. `x=2 pm 8//5`
D. `x=3 pm 16//5`

Answer

To find the equations of the directrices of the hyperbola given by the equation \( 5x^2 - 4y^2 - 30x - 8y - 39 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 5x^2 - 4y^2 - 30x - 8y - 39 = 0 \]
We need to rearrange this equation into the standard form of a hyperbola. First, we group the \(x\) terms and \(y\) terms together.

### Step 2: Complete the square for \(x\) and \(y\)
1. For the \(x\) terms:
\[ 5(x^2 - 6x) = 5((x - 3)^2 - 9) = 5(x - 3)^2 - 45 \]
2. For the \(y\) terms:
\[ -4(y^2 + 2y) = -4((y + 1)^2 - 1) = -4(y + 1)^2 + 4 \]

### Step 3: Substitute back into the equation
Now substitute these completed squares back into the original equation:
\[ 5((x - 3)^2 - 9) - 4((y + 1)^2 - 1) - 39 = 0 \]
This simplifies to:
\[ 5(x - 3)^2 - 45 - 4(y + 1)^2 + 4 - 39 = 0 \]
Combining the constants:
\[ 5(x - 3)^2 - 4(y + 1)^2 - 80 = 0 \]
Rearranging gives:
\[ 5(x - 3)^2 - 4(y + 1)^2 = 80 \]

### Step 4: Divide by 80 to get standard form
\[ \frac{(x - 3)^2}{16} - \frac{(y + 1)^2}{20} = 1 \]
This is now in the standard form of a hyperbola:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
where \(h = 3\), \(k = -1\), \(a^2 = 16\), and \(b^2 = 20\).

### Step 5: Identify \(a\) and \(b\)
From the equation, we find:
\[ a = 4, \quad b = \sqrt{20} = 2\sqrt{5} \]

### Step 6: Calculate the distance to the directrices
The directrices of a hyperbola are given by the equations:
\[ x = h \pm \frac{a^2}{c} \]
where \(c = \sqrt{a^2 + b^2}\).
Calculating \(c\):
\[ c = \sqrt{16 + 20} = \sqrt{36} = 6 \]
Now we can find the directrices:
\[ x = 3 \pm \frac{16}{6} = 3 \pm \frac{8}{3} \]
Calculating the two directrices:
1. \(x = 3 + \frac{8}{3} = \frac{9}{3} + \frac{8}{3} = \frac{17}{3}\)
2. \(x = 3 - \frac{8}{3} = \frac{9}{3} - \frac{8}{3} = \frac{1}{3}\)

### Final Answer
The equations of the directrices of the hyperbola are:
\[ x = \frac{17}{3} \quad \text{and} \quad x = \frac{1}{3} \]