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

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(A) Let the slope of line $L$ be $m$. Since the two given lines are perpendicular to line $L$,they must be parallel to each other.Let the slopes of th

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exactly one value of $p$

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(A) Let the slope of line $L$ be $m$. Since the two given lines are perpendicular to line $L$,they must be parallel to each other.Let the slopes of the two lines be $m_1$ and $m_2$ respectively.For the first line $p(p^2+1)x - y + q = 0$,the slope $m_1 = p(p^2+1)$.For the second line $(p^2+1)^2 x + (p^2+1)y + 2q = 0$,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 $(p^2+1)^2 x + (p^2+1)y + 2q = 0$ are perpendicular to a line $L$ for

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