Question

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. $\text{exactly one value of p}$
• B. $\text{exactly two values of p}$
• C. $\text{more than two values of p}$
• D. $\text{no value of p}$

Answer 1

Answer: (A)

$\text{exactly one value of p}$

If they are perpendicular to a common line then they must be parallel to each other.

$m_1=p(p^2+1)$

$m_2=\frac{{(p^2+1)}^2}{(p^2+1)}=-(p^2+1)$

$m_1=m_2$

$p(p^2+1)=-(p^2+1)$

$\therefore p=-1$

Hence The linesare perpendicular to a common line forexactly one value of p