The line $p(p^2+1)x-y+q=0$ and $ (p^2+1)^2x +(p^2+1)y+2q=0$ are perpendicular to a common line for - Clay6.com, a Free resource for your JEE, AIPMT and Board Exam preparation

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

$\begin{array}{1 1}(A)\;\text {exactly one values of P} \\(B)\;\text{exactly two values of P} \\ (C)\;\text{more than two values of P} \\(D)\;\text{no value of P} \end{array}$