The lines $p\left(p^2+1\right) x-y+q=0$ and $\left(p^2+1\right)^2 x+\left(p^2+1\right) y+2 q=0$ are…

The lines $p\left(p^2+1\right) x-y+q=0$ and $\left(p^2+1\right)^2 x+\left(p^2+1\right) y+2 q=0$ are perpendicular to a common line for
  1. no value of $p$
  2. exactly one value of $p$
  3. exactly two values of $p$
  4. more than two values of $p$

Solution

Lines must be parallel, therefore slopes are equal $\Rightarrow p\left(p^2+1\right)=-\left(p^2+1\right) \Rightarrow p=-1$

Asked in: JEE Main 2009

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