The lines p\left ( p^{2}+1 \right )x-y+q= 0 and

\left ( p^{2}+1 \right )^{2}\! x+\left ( p^{2}+1 \right )y+2q= 0 are perpendicular to a common line for .

  • Option 1)

    exactly one value of  p

  • Option 2)

    exactly two value of  p

  • Option 3)

    more than two value of  p

  • Option 4)

    No value of  p

 

Answers (1)

As we learnt in 

Parallel lines -

\frac{A_{1}}{A_{2}}=\frac{B_{1}}{B_{2}}\neq \frac{C_{1}}{C_{2}}

 

 

- wherein

The two lines are A_{1}x+B_{1}y+C_{1}=0  and A_{2}x+B_{2}y+C_{2}=0

 

 If two lines are perpendicular to same line, they must be parallel.

p(p^{2}+1)x -y+q=0

(p^{2}+1)^{2}x + (p^{2}+1)y+2q=0

\frac{p(p^{2}+1)}{(p^{2}+1)^{2}}=\frac{-1}{(p^{2}+1)}\Rightarrow P=-1

only one value

 


Option 1)

exactly one value of  p

This option is correct.

Option 2)

exactly two value of  p

This option is incorrect.

Option 3)

more than two value of  p

This option is incorrect.

Option 4)

No value of  p

This option is incorrect.

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