The lines `p(p^2+1)x-y+q=0` and `(p^2+1)^2x+(p^2+1)y+2q=0` are perpendicular to a common line for

Options

A. no value of p.
B. exactly one value of p.
C. exactly two values of p.
D. more than two values of p.

Explanation

To solve the problem, we need to find the values of \( p \) for which the two given lines are perpendicular to a common line. The lines are given as:

1. \( p(p^2 + 1)x - y + q = 0 \) (Line 1)
2. \( (p^2 + 1)^2x + (p^2 + 1)y + 2q = 0 \) (Line 2)

### Step 1: Find the slopes of the lines

The slope of a line in the form \( Ax + By + C = 0 \) can be found using the formula:

\[ \text{slope} = -\frac{A}{B} \]

#### Slope of Line 1:

For Line 1, we have:

- \( A = p(p^2 + 1) \)
- \( B = -1 \)

Thus, the slope \( m_1 \) of Line 1 is:

\[ m_1 = -\frac{p(p^2 + 1)}{-1} = p(p^2 + 1) \]

#### Slope of Line 2:

For Line 2, we have:

- \( A = (p^2 + 1)^2 \)
- \( B = (p^2 + 1) \)

Thus, the slope \( m_2 \) of Line 2 is:

\[ m_2 = -\frac{(p^2 + 1)^2}{(p^2 + 1)} = -(p^2 + 1) \]

### Step 2: Set the slopes equal

Since the two lines are perpendicular to a common line, they must be parallel to each other. Therefore, we set the slopes equal:

\[ p(p^2 + 1) = -(p^2 + 1) \]

### Step 3: Solve 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:

1. \( p^2 + 1 = 0 \)
2. \( p + 1 = 0 \)

#### Case 1: \( p^2 + 1 = 0 \)

This case has no real solutions since \( p^2 + 1 \) is always positive for real \( p \).

#### Case 2: \( p + 1 = 0 \)

This gives us:

\[ p = -1 \]

### Conclusion

The only real solution for \( p \) is \( -1 \). Therefore, there is exactly one value of \( p \).

### Final Answer

The correct option is: **Exactly one value of \( p \)**.

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