Question

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

• A. $p=1andq=2or6$
• B. $p=2andq=0or6$
• C. $p=2andq=0or8$
• D. $p=–2andq=–2or8$

Answer 1

The correct option is C

$p=2andq=0or8$

Explanation for the correct option

Finding the condition for which the two lines are mutually perpendicular

Given, the lines are

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

On comparing the above equation with the general form of the equation of pair of straight lines,

$a=2,b=-p,h=2,g=2,f=\frac{q}{2},c=1$

As the given lines are perpendicular,

$a+b=0$

$2-p=0\Rightarrow p=2$

So the equation becomes,

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

Now, we know that the above equation represents a pair of straight lines if,

$$\begin{gathered} abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0 \\ \Rightarrow 2\times (-2)\times 1+2\left(\frac{q}{2}\right)\times 2\times 2-2\left(\frac{q^2}{4}\right)+2\times 2^2-1\times 2^2=0 \\ \Rightarrow -4+4q-\frac{q^2}{2}+8-4=0 \\ \Rightarrow q^2-8q=0 \\ \Rightarrow q(q-8)=0 \\ \Rightarrow q=0,8 \end{gathered}$$

Hence, $p=2$ and $q=0,8$.

Therefore, the correct answer is Option (C).