Question

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 a common line for

• A. No value of $p$
• B. Exactly one value of $p$
• C. Exactly two values of $p$
• D. More than two values of $p$

Answer 1

The correct option is B

Exactly one value of $p$

Solve for $p$

Given,

. $p(p^2+1)x-y+q=0.....\left(1\right)$

${(p^2+1)}^{2}x+(p^2+1)y+2q=0.....\left(2\right)$

The slope of equation $\left(1\right)$ $=p(p^2+1)$

The slope of equation $\left(2\right)$ $=\frac{-({p^2+1)}^2}{p^2+1}=-\left(p^2+1\right)$

Since these lines are perpendicular to a third line, this means they are parallel with each other.

The slope of $\left(1\right)=$The slope of $\left(2\right)$

$$\begin{gathered} \Rightarrow p(p^2+1)=-(p^2+1) \\ \Rightarrow p=-1 \end{gathered}$$

Here, $p$ has only one value, that is, $-1$.

Hence, the correct option is $B$.