If the lines p(p2+1)x−y+q=0 and (p2+1)2x+(p2+1)y+2q=0 are perpendicular to a common line then these lines - must be parallel to each other, ‌∴m1=m2⇒−‌
p(p2+1)
−1
=−‌
(p2+1)2
p2+1
‌⇒(p2+1)(p+1)=0 ‌⇒p=−1 ∴p can have exactly one value.