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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 values of p.

Text Solution

Verified by Experts

The correct Answer is:
B
The lines must be parallel, therefore, slopes are equal. So
`p(p^2+1)=-(p^2+1)rArrp=-1`
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