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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)

best_answer

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

Slope of line 1 = slope of line 2

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

So, we have only one value of p

 

Posted by

rishi.raj

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