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 forA. no value of pB. exactly one value of pC. exactly two values of pD. more than two 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 perpend

1.	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 forA. no value of pB. exactly one value of pC. exactly two values of pD. more than two value of p
Answer» Correct Answer - B Given lines are perpendicular to a common line. So, they are parallel. Therefore, their slopw are equal. i.e., `p(p^(2)+1) = - (p^(2) + 1)rArr p = -1 " "[because p^(2) + 1 ne 0]` Hence, given lines are perpendicular to a common line for excatly one value of p.

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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 forA. no value of pB. exactly one value of pC. exactly two values of pD. more than two value of p

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