# 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

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

## 1 Answer

If the lines $p(p^2+1)x-y+q=0$
$(p^2+1)^2x+(p^2+1)y+2q=0$
are perpendicular to a common line then these lines must be parallel to each other,
$m_1=m_2=> \large\frac{-p(p^2+1)}{-1}=\frac{(p^2+1)^2}{p^2+1}$
$=> (P^2+1)(p+1)=0$
$=> P=-1$
P can have exactly one value.
Hence A is the correct answer.
answered Apr 8, 2014 by

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