JEE MAIN - Mathematics (2009 - No. 15)
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 :
exactly one values of $$p$$
exactly two values of $$p$$
more than two values of $$p$$
no value 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.
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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