The transverse axis of the hyperbola `5x^2-4y^2-30x-8y+121=0` is

Options

A. `y+1=0`
B. `x-3=0`
C. `x+3=0`
D. `y-1=0`

Answer

To find the transverse axis of the hyperbola given by the equation \(5x^2 - 4y^2 - 30x - 8y + 121 = 0\), we will follow these steps:

### Step 1: Rearrange the equation
First, we rearrange the equation to isolate the constant on one side:
\[ 5x^2 - 4y^2 - 30x - 8y = -121 \]

### Step 2: Group the x and y terms
Next, we group the x terms and the y terms:
\[ 5(x^2 - 6x) - 4(y^2 + 2y) = -121 \]

### Step 3: Complete the square for x
To complete the square for the x terms, we take half of the coefficient of \(x\) (which is -6), square it, and add it inside the parentheses:
\[ 5(x^2 - 6x + 9 - 9) - 4(y^2 + 2y) = -121 \]
This simplifies to:
\[ 5((x - 3)^2 - 9) - 4(y^2 + 2y) = -121 \]
Distributing the 5 gives:
\[ 5(x - 3)^2 - 45 - 4(y^2 + 2y) = -121 \]

### Step 4: Complete the square for y
Now, we complete the square for the y terms. We take half of the coefficient of \(y\) (which is 2), square it, and add it:
\[ 5(x - 3)^2 - 45 - 4(y^2 + 2y + 1 - 1) = -121 \]
This simplifies to:
\[ 5(x - 3)^2 - 45 - 4((y + 1)^2 - 1) = -121 \]
Distributing the -4 gives:
\[ 5(x - 3)^2 - 45 - 4(y + 1)^2 + 4 = -121 \]
Combining the constants:
\[ 5(x - 3)^2 - 4(y + 1)^2 - 41 = -121 \]
Adding 41 to both sides:
\[ 5(x - 3)^2 - 4(y + 1)^2 = -80 \]

### Step 5: Divide by -80
To express the equation in standard form, we divide everything by -80:
\[ -\frac{5(x - 3)^2}{80} + \frac{4(y + 1)^2}{80} = 1 \]
This simplifies to:
\[ \frac{(y + 1)^2}{20} - \frac{(x - 3)^2}{16} = 1 \]

### Step 6: Identify the standard form
Now we have the equation in the standard form of a hyperbola:
\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]
where \((h, k) = (3, -1)\), \(a^2 = 20\), and \(b^2 = 16\).

### Step 7: Determine the transverse axis
The transverse axis of a hyperbola of this form is vertical and is given by the line \(x = h\):
\[ x = 3 \]

### Final Answer
The transverse axis of the hyperbola is:
\[ \text{x = 3} \]