The values of `lamda` for which the equation `x^(2)-y^(2)-x-lamday-2=0` represents a pair of straight lines are

Options

A. `3,-3`
B. `-3,1`
C. `3,1`
D. `-1,1`

Answer

To determine the values of \( \lambda \) for which the equation

\[ x^2 - y^2 - x - \lambda y - 2 = 0 \]

represents a pair of straight lines, we can follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation can be rewritten as:

\[ x^2 - y^2 - x - \lambda y - 2 = 0 \]

This can be expressed in the general form of a conic section:

\[ Ax^2 + By^2 + 2Hxy + 2Gx + 2Fy + C = 0 \]

where:
- \( A = 1 \)
- \( B = -1 \)
- \( H = 0 \)
- \( G = -\frac{1}{2} \)
- \( F = -\frac{\lambda}{2} \)
- \( C = -2 \)

### Step 2: Use the condition for a pair of straight lines
For the equation to represent a pair of straight lines, the determinant \( \Delta \) must be equal to zero. The determinant is given by:

\[ \Delta = \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} \]

Substituting the values we have:

\[ \Delta = \begin{vmatrix} 1 & 0 & -\frac{1}{2} \\ 0 & -1 & -\frac{\lambda}{2} \\ -\frac{1}{2} & -\frac{\lambda}{2} & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ \Delta = 1 \cdot \begin{vmatrix} -1 & -\frac{\lambda}{2} \\ -\frac{\lambda}{2} & -2 \end{vmatrix} - 0 + \left(-\frac{1}{2}\right) \cdot \begin{vmatrix} 0 & -1 \\ -\frac{1}{2} & -\frac{\lambda}{2} \end{vmatrix} \]

Calculating the first determinant:

\[ \begin{vmatrix} -1 & -\frac{\lambda}{2} \\ -\frac{\lambda}{2} & -2 \end{vmatrix} = (-1)(-2) - \left(-\frac{\lambda}{2}\right)\left(-\frac{\lambda}{2}\right) = 2 - \frac{\lambda^2}{4} \]

Calculating the second determinant:

\[ \begin{vmatrix} 0 & -1 \\ -\frac{1}{2} & -\frac{\lambda}{2} \end{vmatrix} = 0 \cdot \left(-\frac{\lambda}{2}\right) - (-1)\left(-\frac{1}{2}\right) = -\frac{1}{2} \]

Thus, we have:

\[ \Delta = 1 \cdot \left(2 - \frac{\lambda^2}{4}\right) + \frac{1}{4} = 2 - \frac{\lambda^2}{4} + \frac{1}{4} \]

Combine the terms:

\[ \Delta = 2 + \frac{1}{4} - \frac{\lambda^2}{4} = \frac{8}{4} + \frac{1}{4} - \frac{\lambda^2}{4} = \frac{9 - \lambda^2}{4} \]

### Step 4: Set the determinant to zero
For the equation to represent a pair of straight lines, we set \( \Delta = 0 \):

\[ \frac{9 - \lambda^2}{4} = 0 \]

Multiplying through by 4 gives:

\[ 9 - \lambda^2 = 0 \]

### Step 5: Solve for \( \lambda \)
Rearranging gives:

\[ \lambda^2 = 9 \]

Taking the square root of both sides yields:

\[ \lambda = \pm 3 \]

### Conclusion
The values of \( \lambda \) for which the equation represents a pair of straight lines are:

\[ \lambda = 3 \quad \text{and} \quad \lambda = -3 \]