Question

The equation $2x^2+4xy-p{y}^2+4x+qy+1=0$ will represent two mutually perpendicular straight lines if ?

• A. $p=1\text{and}q=2\text{or}6$
• B. $p=-2\text{and}q=-2\text{or}8$
• C. $p=2\text{and}q=0\text{or}8$
• D. $p=2\text{and}q=0\text{or}6$

Answer 1

Answer: (C)

$p=2\text{and}q=0\text{or}8$

$2x^2+4xy-p{y}^2+4x+qy+1=0$

$2-p=0$ [Coefficient of $(x^2+y^2)=0]$

$p=2$

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

[Condition for the equation to represent straight lines]

$2(-2)\left(1\right)+2\left(\frac{q}{2}\right)\left(2\right)\left(2\right)-2{\left(\frac{q}{2}\right)}^2+2(2{)}^2-1(2{)}^2=0$

$q^2-8q=0$

$q(q-8)=0q=0$ or $8$.