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

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

A

no value of p

B

exactly one value of p

C

exactly two values of p

D

more than two value of p

Text Solution

Verified by Experts

The correct Answer is:
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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