The lines $p\left( p^{2} + 1 \right)x - y + q = 0$ and $(p^{2} + {1)}^{2}x + \left( p^{2} + 1 \right)y + 2q = 0$ are perpendicular to a common line for

Question

Dr. Abubacker · Accepted Answer

(a)
Line perpendicular to same line are parallel to each other.
$	herefore\  - p\left( p^{2} + 1 ight) = p^{2} + 1$
$\Rightarrow \ p = - 1$
$	herefore$ There is exactly one value of $p$.

Q. The lines $p\left( p^{2} + 1 \right)x - y + q = 0$ and $(p^{2} + {1)}^{2}x + \left( p^{2} + 1 \right)y + 2q = 0$ are perpendicular to a common line for

Explanation

Reviewed by Dr. Abubacker

Want more practice?