The lines $p\left(p^2+1\right) x-y+q=0$ and $\left(p^2+1\right)^2 x+\left(p^2+1\right)$ $y+2 q=0$ are…
The lines $p\left(p^2+1\right) x-y+q=0$ and $\left(p^2+1\right)^2 x+\left(p^2+1\right)$ $y+2 q=0$ are perpendicular to a line $L$ for
exactly one value of $p$
exactly two values of $p$
more than two values of $p$
no value of $p$
Solution
For 1st line $m_1=p\left(p^2+1\right)$ for $2 \mathrm{nd}$ line $\mathrm{m}_2=-\left(\mathrm{p}^2+1\right)$
Since, if these two line are perpendicular to the line $\mathrm{L}$ then
$\begin{aligned}
& \mathrm{m}_1=\mathrm{m}_2 \\
& \Rightarrow p\left(p^2+1\right)=-\left(p^2+1\right) \Rightarrow p=-1
\end{aligned}$