The lines p(p2+1)x−y+q=0 and (p2+1)2x+(p2+1)y+2q=0 are perpendicular to a common line for:
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Official solution
If the lines \[p\left( {{p^2} + 1} \right)x - y + q = 0\]
and \[{\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0\]
are perpendicular to a common line then these lines -
must be parallel to each other,
\[\therefore\] \[{m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} + 1} \right)} \over { - 1}} = - {{{{\left( {{p^2} + 1} \right)}^2}} \over {{p^2} + 1}}\]
\[ \Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0\]
\[ \Rightarrow p = - 1\]
\[\therefore\] p can have exactly one value.