For What value of `lamda` the equation of conic 2xy+4x-6y+`lamda`=0 represents two intersecting straight lines, if `lamda=17`, then this equation represents?

Answer

To solve the problem step by step, we will first determine the value of `lambda` for which the given conic represents two intersecting straight lines. Then, we will analyze the case when `lambda = 17`.

### Step 1: Identify the coefficients in the conic equation
The given conic equation is:
\[ 2xy + 4x - 6y + \lambda = 0 \]

We can rewrite it in the standard form of a conic:
\[ Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \]
where:
- \( A = 0 \) (no \( x^2 \) term)
- \( B = 0 \) (no \( y^2 \) term)
- \( H = 1 \) (coefficient of \( xy \))
- \( G = 2 \) (coefficient of \( x \))
- \( F = -3 \) (coefficient of \( y \))
- \( C = \lambda \)

### Step 2: Write the discriminant condition for the conic
For the conic to represent two intersecting straight lines, the discriminant must be equal to zero:
\[ D = 2H^2 - 4AB = 0 \]

Substituting the values we have:
- \( A = 0 \)
- \( B = 0 \)
- \( H = 1 \)

So,
\[ D = 2(1)^2 - 4(0)(0) = 2 \]

Since \( D \) is not equal to zero, we need to use the determinant method for the general conic:
\[ \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = 0 \]

### Step 3: Set up the determinant
Substituting the values:
\[ \begin{vmatrix} 0 & 1 & 2 \\ 1 & 0 & -3 \\ 2 & -3 & \lambda \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ = 0 \cdot (0 \cdot \lambda - (-3) \cdot (-3)) - 1 \cdot (1 \cdot \lambda - (-3) \cdot 2) + 2 \cdot (1 \cdot (-3) - 0 \cdot 2) \]
\[ = 0 - (\lambda + 6) - 6 \]
\[ = -\lambda - 12 \]

Setting the determinant to zero for the conic to represent two intersecting lines:
\[ -\lambda - 12 = 0 \implies \lambda = -12 \]

### Step 5: Analyze the case when `lambda = 17`
Now, substituting \( \lambda = 17 \) into the conic equation:
\[ 2xy + 4x - 6y + 17 = 0 \]

### Step 6: Determine the type of conic for `lambda = 17`
We check the condition:
\[ 2H^2 - 4AB \]
Substituting \( A = 0, B = 0, H = 1 \):
\[ D = 2(1)^2 - 4(0)(0) = 2 \]

Since \( D > 0 \), the conic represents a hyperbola.

### Final Answer
Thus, the value of \( \lambda \) for which the equation represents two intersecting straight lines is \( \lambda = -12 \). When \( \lambda = 17 \), the equation represents a hyperbola.