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The lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$ and $$\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$$ $$=0$$ are perpendicular to a common line for :

JEE · Math · previous-year question

  1. A.exactly one values of $$p$$correct
  2. B.exactly two values of $$p$$
  3. C.more than two values of $$p$$
  4. D.no value of $$p$$

Answer

A. exactly one values of $$p$$

Explanation

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.

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