Area enclosed by the pair of lines $S\equiv a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ and $S\text{'}\equiv a{x}^2+2hxy+b{y}^2-2gx-2fy+c=0$

$S=\left(l_{1}x+m_{1}y+n_1\right)\left(l_{2}x+m_{2}y+n_2\right)$ replace $x=-x,y=-y$

$S\text{'}=(-l_{1}x-m_{1}y+n_1)(-l_{2}x-m_{2}y+n_1)$

above four lines form a parallelogram 

area $=\left|\frac{{\displaystyle (c_1-c_2)(d_1-d_2)}}{{\displaystyle a_1{b}_2-a_2{b}_1}}\right|$

$=\frac{\left|\left(2n_1\right)\left(2n_2\right)\right|}{\sqrt{{(l_1{m}_2-l_2{m}_1)}^2}}=\frac{\left|4n_1{n}_2\right|}{\sqrt{{(l_1{m}_2+l_2{m}_1)}^2-4l_1{l}_2{m}_1{m}_2}}$

$=\frac{4\left|c\right|}{2\sqrt{h^2-ab}}=\frac{2\left|c\right|}{\sqrt{h^2-ab}}$