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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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Knowledge Check

  • 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 common line for

    A
    exactly one value of p
    B
    exactly two value of p
    C
    more than two values of p
    D
    no 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 common line for:

    A
    more than two value of p
    B
    no value of p
    C
    exactly one value of p
    D
    exactly two values of p

  • If the lines (p-q)x^2+2(p+q)xy+(q-p)y^2=0 are mutually perpendicular , then

    A
    `p=q`
    B
    `p=0`
    C
    `q=0`
    D
    p and q may have any value

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