To find the vertex of the conic section given by the equation \(16x^2 - y^2 - 32x - 6y - 57 = 0\), we will follow these steps:
### Step 1: Rearrange the equation
First, we rearrange the equation to isolate the constant on one side:
\[
16x^2 - y^2 - 32x - 6y = 57
\]
### Step 2: Group the \(x\) and \(y\) terms
Next, we group the \(x\) terms and the \(y\) terms:
\[
16x^2 - 32x - y^2 - 6y = 57
\]
### Step 3: Complete the square for \(x\)
To complete the square for the \(x\) terms, we take the coefficient of \(x\), which is \(-32\), halve it to get \(-16\), and then square it to get \(256\). We add and subtract \(256\):
\[
16(x^2 - 2x + 16) - y^2 - 6y = 57 + 256
\]
This simplifies to:
\[
16(x - 1)^2 - y^2 - 6y = 313
\]
### Step 4: Complete the square for \(y\)
Now, we complete the square for the \(y\) terms. The coefficient of \(y\) is \(-6\), halve it to get \(-3\), and square it to get \(9\). We add and subtract \(9\):
\[
16(x - 1)^2 - (y^2 + 6y + 9) = 313 + 9
\]
This simplifies to:
\[
16(x - 1)^2 - (y + 3)^2 = 322
\]
### Step 5: Divide by 322
Now, we divide the entire equation by \(322\) to get it into standard form:
\[
\frac{16(x - 1)^2}{322} - \frac{(y + 3)^2}{322} = 1
\]
### Step 6: Identify the standard form parameters
This equation can be rewritten as:
\[
\frac{(x - 1)^2}{\frac{322}{16}} - \frac{(y + 3)^2}{322} = 1
\]
From this, we identify:
- \(h = 1\)
- \(k = -3\)
- \(a = \sqrt{\frac{322}{16}} = \frac{\sqrt{322}}{4}\)
- \(b = \sqrt{322}\)
### Step 7: Find the vertices
The vertices of a hyperbola oriented along the y-axis are given by:
\[
(h, k \pm a)
\]
Calculating the vertices:
1. \( (h, k + a) = (1, -3 + \frac{\sqrt{322}}{4}) \)
2. \( (h, k - a) = (1, -3 - \frac{\sqrt{322}}{4}) \)
### Step 8: Compare with options
Now we check the options provided to see which vertex matches.
### Final Answer
The vertices calculated are:
1. \( (1, -3 + \frac{\sqrt{322}}{4}) \)
2. \( (1, -3 - \frac{\sqrt{322}}{4}) \)
If the options include \( (-1, -3) \) or \( (3, -3) \), we can check which matches our calculations.
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