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 $p$

If two lines are perpendicular to a common line then they are mutually parallel in 2D.
Hence
$p(p^2+1)x-y+q=0$ is parallel to $(p^2+1{)}^{2}x+(p^2+1)y+2q=0$
Hence, slopes are equal
$p(p^2+1)=\frac{-1(p^2+1{)}^2}{p^2+1}$
Thus
$\frac{p(p^2+1)}{(p^2+1{)}^2}=\frac{-1}{p^2+1}$
$\frac{p}{p^2+1}=\frac{-1}{p^2+1}$
Hence
$p=-1$
Thus, there exists exactly one value of $p$.