The equation `x^(2)+4xy+4y^(2)-3x-6y-4=0` represents a

Options

A. circle
B. parabola
C. a pair of straght lines
D. none of these

Explanation

To determine the type of conic section represented by the equation \( x^2 + 4xy + 4y^2 - 3x - 6y - 4 = 0 \), we will follow these steps:

### Step 1: Identify coefficients
The general form of a conic section is given by:
\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
From the given equation, we can identify the coefficients:
- \( A = 1 \)
- \( B = 4 \)
- \( C = 4 \)
- \( D = -3 \)
- \( E = -6 \)
- \( F = -4 \)

### Step 2: Calculate the discriminant (delta)
The discriminant \( \Delta \) is calculated using the formula:
\[ \Delta = ABC + 2FGH - AF^2 - BG^2 - CE^2 \]
where \( G = \frac{D}{2} \), \( H = \frac{E}{2} \), and \( F = F \).

First, we calculate \( G \) and \( H \):
- \( G = \frac{-3}{2} = -\frac{3}{2} \)
- \( H = \frac{-6}{2} = -3 \)

Now substituting into the discriminant formula:
\[ \Delta = (1)(4)(-4) + 2(-3)(-3)(2) - (1)(-3)^2 - (4)(-\frac{3}{2})^2 - (4)(-3)^2 \]
Calculating each term:
1. \( ABC = 1 \cdot 4 \cdot -4 = -16 \)
2. \( 2FGH = 2 \cdot -3 \cdot -3 \cdot 2 = 36 \)
3. \( AF^2 = 1 \cdot (-3)^2 = 9 \)
4. \( BG^2 = 4 \cdot \left(-\frac{3}{2}\right)^2 = 4 \cdot \frac{9}{4} = 9 \)
5. \( CE^2 = 4 \cdot (-3)^2 = 4 \cdot 9 = 36 \)

Now substituting these values back into the discriminant:
\[ \Delta = -16 + 36 - 9 - 9 - 36 \]
Calculating:
\[ \Delta = -16 + 36 - 54 = -34 \]

### Step 3: Check conditions for conic sections
1. **Circle**: For a circle, \( B = 0 \) and \( \Delta > 0 \). (Not applicable here)
2. **Parabola**: For a parabola, \( \Delta = 0 \). (Not applicable here)
3. **Pair of straight lines**: For a pair of straight lines, \( \Delta = 0 \) and \( H^2 - AB = 0 \).

### Step 4: Calculate \( H^2 - AB \)
Now we calculate \( H^2 - AB \):
\[ H^2 = (-3)^2 = 9 \]
\[ AB = 1 \cdot 4 = 4 \]
Thus,
\[ H^2 - AB = 9 - 4 = 5 \]

### Conclusion
Since \( \Delta = 0 \) and \( H^2 - AB = 5 \neq 0 \), we conclude that the equation represents a pair of straight lines.