The vertices of the hyperbola `x^(2)-3y^(2)+2x+12y+1=0` are

Options

A. `(pm3,0)`
B. `(1 pm 2,2)`
C. `(-1,2 pm 2)`
D. `(1, -2 pm3)`

Answer

To find the vertices of the hyperbola given by the equation \( x^2 - 3y^2 + 2x + 12y + 1 = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
Start by rearranging the equation to group the \(x\) and \(y\) terms together:
\[ x^2 + 2x - 3y^2 + 12y + 1 = 0 \]

### Step 2: Complete the square for \(x\)
To complete the square for the \(x\) terms \(x^2 + 2x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]

### Step 3: Complete the square for \(y\)
Now, for the \(y\) terms \(-3y^2 + 12y\), factor out \(-3\):
\[ -3(y^2 - 4y) = -3((y - 2)^2 - 4) = -3(y - 2)^2 + 12 \]

### Step 4: Substitute back into the equation
Substituting the completed squares back into the equation gives:
\[ ((x + 1)^2 - 1) - 3((y - 2)^2 - 4) + 1 = 0 \]
This simplifies to:
\[ (x + 1)^2 - 1 - 3(y - 2)^2 + 12 + 1 = 0 \]
\[ (x + 1)^2 - 3(y - 2)^2 + 12 = 0 \]
Rearranging gives:
\[ (x + 1)^2 - 3(y - 2)^2 = -12 \]

### Step 5: Divide by -12
To put this into standard form, divide the entire equation by -12:
\[ -\frac{(x + 1)^2}{12} + \frac{(y - 2)^2}{4} = 1 \]
This can be rewritten as:
\[ \frac{(y - 2)^2}{4} - \frac{(x + 1)^2}{12} = 1 \]

### Step 6: Identify the parameters
Now, we can compare this equation with the standard form of a hyperbola:
\[ \frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1 \]
From our equation, we identify:
- \(k = 2\)
- \(b^2 = 4 \implies b = 2\)
- \(h = -1\)
- \(a^2 = 12 \implies a = \sqrt{12} = 2\sqrt{3}\)

### Step 7: Find the vertices
The vertices of the hyperbola are given by the coordinates:
\[ (h, k \pm b) = (-1, 2 \pm 2) \]
Calculating these gives:
- \( (-1, 2 + 2) = (-1, 4) \)
- \( (-1, 2 - 2) = (-1, 0) \)

### Final Answer
Thus, the vertices of the hyperbola are:
\[ (-1, 4) \text{ and } (-1, 0) \]

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