it is known that the equation of hyperbola and that of its pair of asymptotes differ by constant . If equation of hyperbola is `x^(2) +4xy+3y^(2) -4x+2y+1=0` then find the equation of its pair of asymptoes.

Explanation

To find the equation of the pair of asymptotes for the given hyperbola, we will follow these steps:

### Step 1: Write down the equation of the hyperbola
The equation of the hyperbola is given as:
\[ x^2 + 4xy + 3y^2 - 4x + 2y + 1 = 0 \]

### Step 2: Assume the equation of the pair of asymptotes
We know that the equation of the hyperbola and that of its pair of asymptotes differ by a constant. Therefore, we can express the equation of the pair of asymptotes as:
\[ x^2 + 4xy + 3y^2 - 4x + 2y + 1 + k = 0 \]
where \( k \) is a constant.

### Step 3: Set up the determinant condition for the pair of straight lines
For the equation to represent a pair of straight lines, the determinant of the coefficients must equal zero. The coefficients of the quadratic terms can be arranged in a matrix as follows:
\[ \begin{vmatrix} 1 & 2 & -2 \\ 2 & 3 & 1 \\ -2 & 1 & 1+k \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ \begin{vmatrix} 1 & 2 & -2 \\ 2 & 3 & 1 \\ -2 & 1 & 1+k \end{vmatrix} \]
Using the determinant formula for a 3x3 matrix:
\[ = 1 \cdot \begin{vmatrix} 3 & 1 \\ 1 & 1+k \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ -2 & 1+k \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 3 & 1 \\ 1 & 1+k \end{vmatrix} = 3(1+k) - 1(1) = 3 + 3k - 1 = 2 + 3k \)
2. \( \begin{vmatrix} 2 & 1 \\ -2 & 1+k \end{vmatrix} = 2(1+k) - 1(-2) = 2 + 2k + 2 = 4 + 2k \)
3. \( \begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} = 2(1) - 3(-2) = 2 + 6 = 8 \)

Substituting back into the determinant:
\[ = 1(2 + 3k) - 2(4 + 2k) - 2(8) \]
\[ = 2 + 3k - 8 - 4k - 16 \]
\[ = -22 - k = 0 \]

### Step 5: Solve for \( k \)
From the equation \( -22 - k = 0 \), we find:
\[ k = -22 \]

### Step 6: Write the equation of the asymptotes
Substituting \( k \) back into the equation of the asymptotes:
\[ x^2 + 4xy + 3y^2 - 4x + 2y + 1 - 22 = 0 \]
This simplifies to:
\[ x^2 + 4xy + 3y^2 - 4x + 2y - 21 = 0 \]

### Final Answer
The equation of the pair of asymptotes is:
\[ x^2 + 4xy + 3y^2 - 4x + 2y - 21 = 0 \]