The lines p(p^2 + 1)x - y + q = 0 and (p^2 + | vedclass

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(D) If two lines are perpendicular to a common line,they must be parallel to each other.The slope of the first line $m_1$ is given by $m_1 = p(p^2 + 1

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For exactly one value of $p$

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(D) If two lines are perpendicular to a common line,they must be parallel to each other.The slope of the first line $m_1$ is given by $m_1 = p(p^2 + 1)$.The slope of the second line $m_2$ is given by $m_2 = -\frac{(p^2 + 1)^2}{p^2 + 1} = -(p^2 + 1)$.For the lines to be parallel,$m_1 = m_2$.$p(p^2 + 1) = -(p^2 + 1)$.Since $p^2 + 1 
eq 0$ for any real $p$,we can divide both sides by $(p^2 + 1)$:$p = -1$.Thus,the lines are perpendicular to a common line for exactly one value of $p$.

The lines $p(p^2 + 1)x - y + q = 0$ and $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$ are perpendicular to a common line for which value of $p$?

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