The equation `2x^2+4xy-py^2+4x+qy+1=0` will represent two mutually perpendicular straight lines , if

Options

A. `p=1 and q=1 or 6`
B. `p=2 and q=0 or 6`
C. `p=2 and q =0 or 8`
D. `p=-2 and 1=-2 or 8`

Explanation

To determine the conditions under which the equation \(2x^2 + 4xy - py^2 + 4x + qy + 1 = 0\) represents two mutually perpendicular straight lines, we can follow these steps:

### Step 1: Identify the coefficients
The general form of the equation of the conic is given by:
\[ Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \]
From the given equation, we can identify the coefficients:
- \(A = 2\)
- \(H = 2\)
- \(B = -p\)
- \(G = 2\)
- \(F = \frac{q}{2}\)
- \(C = 1\)

### Step 2: Condition for perpendicular lines
For the equation to represent two mutually perpendicular straight lines, the condition is:
\[ A + B = 0 \]
Substituting the values we have:
\[ 2 - p = 0 \]
From this, we can solve for \(p\):
\[ p = 2 \]

### Step 3: Condition for the equation to represent straight lines
Next, we need to ensure that the determinant condition for the conic section is satisfied. The determinant for the conic is given by:
\[ \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = 0 \]
Substituting the values we have:
\[ \begin{vmatrix} 2 & 2 & 2 \\ 2 & -p & \frac{q}{2} \\ 2 & \frac{q}{2} & 1 \end{vmatrix} = 0 \]
Substituting \(p = 2\):
\[ \begin{vmatrix} 2 & 2 & 2 \\ 2 & -2 & \frac{q}{2} \\ 2 & \frac{q}{2} & 1 \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ = 2 \begin{vmatrix} -2 & \frac{q}{2} \\ \frac{q}{2} & 1 \end{vmatrix} - 2 \begin{vmatrix} 2 & \frac{q}{2} \\ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} 2 & -2 \\ 2 & \frac{q}{2} \end{vmatrix} \]
Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} -2 & \frac{q}{2} \\ \frac{q}{2} & 1 \end{vmatrix} = -2 \cdot 1 - \frac{q}{2} \cdot \frac{q}{2} = -2 - \frac{q^2}{4} \)

2. \( \begin{vmatrix} 2 & \frac{q}{2} \\ 2 & 1 \end{vmatrix} = 2 \cdot 1 - 2 \cdot \frac{q}{2} = 2 - q \)

3. \( \begin{vmatrix} 2 & -2 \\ 2 & \frac{q}{2} \end{vmatrix} = 2 \cdot \frac{q}{2} - 2 \cdot -2 = q + 4 \)

Substituting back into the determinant:
\[ 2(-2 - \frac{q^2}{4}) - 2(2 - q) + 2(q + 4) = 0 \]
This simplifies to:
\[ -4 - \frac{q^2}{2} - 4 + 2q + 2q + 8 = 0 \]
Combining like terms:
\[ - \frac{q^2}{2} + 4q = 0 \]
Factoring out \(q\):
\[ q(-\frac{q}{2} + 4) = 0 \]
Thus, \(q = 0\) or \(-\frac{q}{2} + 4 = 0\) gives:
\[ q = 8 \]

### Final Result
The values of \(p\) and \(q\) for the equation to represent two mutually perpendicular straight lines are:
\[ p = 2, \quad q = 0 \text{ or } q = 8 \]