The equation `ax^2+4xy+y^2+ax+3y+2=0` represents a parabola. Find the value of `a`.

Answer

To solve the problem, we need to find the value of \( a \) such that the equation \( ax^2 + 4xy + y^2 + ax + 3y + 2 = 0 \) represents a parabola.

### Step-by-Step Solution:

1. **Identify the general form of the conic section**:
The general equation of a conic section is given by:
\[ Ax^2 + 2Hxy + By^2 + 2Fx + 2Gy + C = 0 \]
Here, we can identify:
- \( A = a \)
- \( 2H = 4 \) (which means \( H = 2 \))
- \( B = 1 \)
- \( F = \frac{a}{2} \)
- \( G = \frac{3}{2} \)
- \( C = 2 \)

2. **Use the condition for a parabola**:
For the equation to represent a parabola, the following condition must hold:
\[ H^2 - AB = 0 \]
Substituting the values we have:
\[ H^2 - AB = 2^2 - (a)(1) = 0 \]

3. **Solve for \( a \)**:
Now, simplifying the equation:
\[ 4 - a = 0 \]
This leads us to:
\[ a = 4 \]

### Conclusion:
Thus, the value of \( a \) is \( 4 \).

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