If the conic $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines then it must be of the form $(lx+my+n)(px+qy+r)=0$, where (comparing coefficients) $a = lp$, $h = \frac12(lq+mp)$, $b = mq$, ..., $c = nr.$ The conic $ax^2+2hxy+by^2-2gx-2fy+c=0$ is then given by $(lx+my-n)(px+qy-r)=0.$If each of the equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ and $ax^2+2hxy+by^2-2gx-2fy+c=0$ represents a pair of straight lines, find the area of the parallelogram enclosed by them.
I think you are misunderstanding the question. Each equation represents a pair of lines. For example [tex]a^2x^2- b^2y^2= (ax- by)(ax+by)= 0[/tex] gives the lines ax- by= 0 and ax+ by= 0.You should note that if these are lines, none of the terms $x^2, xy$ and $y^2$ should appear. This should tell you something about their respective coefficients.