Question

Find $p$ and $q$, If the equation $2x^2+4xy-p{y}^2+4x+qy+1=0$. represents a pair of perpendicular line.

Answer 1

Given equation is

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

comparing it with

$a{x}^2+2hxy+h{y}^2+2gx+2fy+c=0,$ we get

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

Since, The given equation represents a

pair of line perpendicular to each other.

now, $a+b=0$

$2+(-p)=0$

$p=2$

Also, The equation represents a pair of lines

$\therefore \left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

$\therefore \left|\begin{array}{ccc}2& 2& 2\\ 2& -2& \frac{q}{2}\\ 2& \frac{q}{2}& 1\end{array}\right|=0$

$\therefore 2(-2\times 1-\frac{q}{2}\times \frac{q}{2})-2(2\times 1-2\times \frac{q}{2})+2(2\times \frac{q}{2}-(-2)\times 2)=0$

$\therefore 2(-2-\frac{q^2}{4})-2(2-q)+2(q+4)=0$

$\therefore 2[-2-\frac{q^2}{4}-2+q+q+4]=0$

$\therefore [-4-\frac{q^2}{4}+2q+4]=0$

$\therefore [2q-\frac{q^2}{4}]=0$

$\therefore 2q=\frac{q^2}{4}$

$\therefore q^2=8q$

$\therefore q^2-8q=0$

$\therefore q^2-8q=0$

$\therefore q(q-8)=0$

$\therefore q=0,q=8$