Find the lengths of transverse axis of the hyperbola `16x^(2)-3y^(2)-32x-12y-44=0`

Answer

To find the lengths of the transverse axis of the hyperbola given by the equation \( 16x^2 - 3y^2 - 32x - 12y - 44 = 0 \), we will follow these steps:

### Step 1: Rearranging the equation
We start with the equation:
\[ 16x^2 - 3y^2 - 32x - 12y - 44 = 0 \]
Rearranging gives:
\[ 16x^2 - 32x - 3y^2 - 12y = 44 \]

### Step 2: Completing the square for \(x\)
For the \(x\) terms:
\[ 16(x^2 - 2x) \]
To complete the square, we take half of the coefficient of \(x\) (which is \(-2\)), square it, and add/subtract it inside the bracket:
\[ 16\left((x - 1)^2 - 1\right) = 16(x - 1)^2 - 16 \]

### Step 3: Completing the square for \(y\)
For the \(y\) terms:
\[ -3(y^2 + 4y) \]
Completing the square gives:
\[ -3\left((y + 2)^2 - 4\right) = -3(y + 2)^2 + 12 \]

### Step 4: Substitute back into the equation
Now substituting back into the equation:
\[ 16(x - 1)^2 - 16 - 3(y + 2)^2 + 12 = 44 \]
This simplifies to:
\[ 16(x - 1)^2 - 3(y + 2)^2 - 4 = 44 \]
Adding 4 to both sides:
\[ 16(x - 1)^2 - 3(y + 2)^2 = 48 \]

### Step 5: Divide by 48
Now, divide the entire equation by 48 to get it into standard form:
\[ \frac{16(x - 1)^2}{48} - \frac{3(y + 2)^2}{48} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{3} - \frac{(y + 2)^2}{16} = 1 \]

### Step 6: Identify \(a\) and \(b\)
From the standard form of the hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), we can identify:
- \(a^2 = 3\) so \(a = \sqrt{3}\)
- \(b^2 = 16\) so \(b = 4\)

### Step 7: Calculate the length of the transverse axis
The length of the transverse axis is given by:
\[ \text{Length of Transverse Axis} = 2a = 2\sqrt{3} \]

Thus, the length of the transverse axis of the hyperbola is \(2\sqrt{3}\).

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