Question

Find the condition that one of the lines given by the equation $a{x}^2+2hxy+b{y}^2=0$ be perpendicular to one of those given by $d{x}^2+2h^{\prime }xy+b^{\prime }{y}^2=0$.

Answer 1

If one of lines given by the first pair by $y=mx$, then
by the given condition one of the lines given by the $2$nd pair should be
$y=-\left(1/3\right)x$.
$\therefore b{m}^2+2hm+a=0$
and $b(-1/m{)}^2+2h(-1/m)+d=0$
or $d{m}^2-2hm+b=0$
Ans. $(ad-bb{)}^2-4(hb+ha\left)\right(bh+hd)=0$.