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

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(A) If two lines are perpendicular to a common line,they must be parallel to each other.Let the slopes of the lines be $m_1$ and $m_2$.For the first l

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exactly one value of $p$

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(A) If two lines are perpendicular to a common line,they must be parallel to each other.Let the slopes of the lines be $m_1$ and $m_2$.For the first line $p(p^2 + 1)x - y + q = 0$,the slope $m_1 = \frac{p(p^2 + 1)}{1} = p(p^2 + 1)$.For the second line $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$,the slope $m_2 = -\frac{(p^2 + 1)^2}{p^2 + 1} = -(p^2 + 1)$.Since the lines are 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 by $(p^2 + 1)$.$p = -1$.Thus,there is 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 :

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