If the equation `x^2-y^2-2x+2y+lamda=0` represent a degenerate conic . Find the value of `lamda`.

Answer

To find the value of \(\lambda\) such that the equation \(x^2 - y^2 - 2x + 2y + \lambda = 0\) represents a degenerate conic, we will follow these steps:

### Step 1: Identify the coefficients from the conic equation
The general form of a conic section is given by:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]
From the given equation \(x^2 - y^2 - 2x + 2y + \lambda = 0\), we can identify the coefficients:
- \(a = 1\)
- \(h = 0\)
- \(b = -1\)
- \(g = -1\)
- \(f = 1\)
- \(c = \lambda\)

### Step 2: Write the discriminant condition for a degenerate conic
For a conic to be degenerate, its discriminant \(D\) must equal zero. The discriminant \(D\) is given by the determinant of the following matrix:
\[ D = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \]
Substituting the values we identified:
\[ D = \begin{vmatrix} 1 & 0 & -1 \\ 0 & -1 & 1 \\ -1 & 1 & \lambda \end{vmatrix} \]

### Step 3: Calculate the determinant
Now we will calculate the determinant \(D\):
\[ D = 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & \lambda \end{vmatrix} - 0 + (-1) \cdot \begin{vmatrix} 0 & -1 \\ -1 & 1 \end{vmatrix} \]
Calculating the first determinant:
\[ \begin{vmatrix} -1 & 1 \\ 1 & \lambda \end{vmatrix} = (-1) \cdot \lambda - (1 \cdot 1) = -\lambda - 1 \]
Calculating the second determinant:
\[ \begin{vmatrix} 0 & -1 \\ -1 & 1 \end{vmatrix} = (0 \cdot 1) - (-1 \cdot -1) = 0 - 1 = -1 \]
Thus,
\[ D = 1(-\lambda - 1) + 1 = -\lambda - 1 + 1 = -\lambda \]

### Step 4: Set the discriminant equal to zero
For the conic to be degenerate, we set the discriminant equal to zero:
\[ -\lambda = 0 \]
This leads to:
\[ \lambda = 0 \]

### Step 5: Conclusion
The value of \(\lambda\) for which the equation represents a degenerate conic is:
\[ \lambda = 0 \]