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 line L for

Question

Accepted Answer

Correct answer: A. (a) For 1st line m_1=p(p^2+1) for 2 nd line m_2=-(p^2+1) Since, if these two line are perpendicular to the line L then m_1=m_2 ⇒p(p^2+1)=-(p^2+1) ⇒p=-1

$The lines p (p^{2} + 1) x - y + q = 0 and (p^{2} + 1)^{2} x + (p^{2} + 1) y + 2 q = 0 are perpendicular to a line L for$

Solution

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