If the lines `p(p^(2)+1)x-y+q=0` and `(p^(2)+1)^(2)x+(p^(2)+1)y+2q=0` are perpendicular to the same line, then the value of p is

Options

A. (a) 1
B. (b) `-1`
C. (c) 2
D. (d) `-2`

Answer

To solve the problem, we need to find the value of \( p \) such that the two given lines are perpendicular to the same line, which implies that they are parallel to each other.

### Step-by-step Solution:

1. **Identify the equations of the lines:**
The first line \( L_1 \) is given by:
\[ p(p^2 + 1)x - y + q = 0 \]
The second line \( L_2 \) is given by:
\[ (p^2 + 1)^2 x + (p^2 + 1)y + 2q = 0 \]

2. **Rearranging the equations to find slopes:**
We need to express both lines in the slope-intercept form \( y = mx + c \).

- For line \( L_1 \):
\[ y = p(p^2 + 1)x + q \]
The slope \( m_1 \) of line \( L_1 \) is:
\[ m_1 = p(p^2 + 1) \]

- For line \( L_2 \):
Rearranging gives:
\[ (p^2 + 1)y = -(p^2 + 1)^2 x - 2q \]
Dividing by \( (p^2 + 1) \) (assuming \( p^2 + 1 \neq 0 \)):
\[ y = -\frac{(p^2 + 1)^2}{(p^2 + 1)}x - \frac{2q}{(p^2 + 1)} \]
Simplifying gives:
\[ y = -(p^2 + 1)x - \frac{2q}{(p^2 + 1)} \]
The slope \( m_2 \) of line \( L_2 \) is:
\[ m_2 = -(p^2 + 1) \]

3. **Setting the slopes equal:**
Since the two lines are parallel, we set their slopes equal:
\[ p(p^2 + 1) = -(p^2 + 1) \]

4. **Solving for \( p \):**
Rearranging the equation gives:
\[ p(p^2 + 1) + (p^2 + 1) = 0 \]
Factoring out \( (p^2 + 1) \):
\[ (p^2 + 1)(p + 1) = 0 \]
This gives us two cases:
- \( p^2 + 1 = 0 \) (not possible for real \( p \))
- \( p + 1 = 0 \) which gives \( p = -1 \)

5. **Conclusion:**
Thus, the value of \( p \) is:
\[ \boxed{-1} \]