QJEEM09Ma809b: JEE Main 2009 Mathematics - Class 11 JEEM09Ma809b Solution: The lines p ( p 2 + 1 ) x − y + q = 0 and ( p 2 + 1 ) 2 x + ( p …

Question

Verified Solution • 2025–26 Session moderate

The lines

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

and

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

= 0

are perpendicular to a common line for :

Accepted Answer

If the lines

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

and

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

are perpendicular to a common line then these lines -

must be parallel to each other,

∴

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

\Rightarrow (p^{2} + 1) (p + 1) = 0

\Rightarrow p = - 1

∴

p

can have exactly one value.

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