If the equation S$\equiv$ax²+2hxy+by²+2gx+2fy+c=0 represents a pair of parallel straight lines, then show that (i) h² = ab (ii) af² = bg² and (iii) the distance between the parallel lines is $2\sqrt{\frac{{\mathrm{g}}^2-\mathrm{ac}}{\mathrm{a}(\mathrm{a}+\mathrm{b})}}=2\sqrt{\frac{{\mathrm{f}}^2-\mathrm{bc}}{\mathrm{b}(\mathrm{a}+\mathrm{b})}}.$

Let the parallel lines represented by S = 0 be $\mathrm{lx}+\mathrm{my}+{\mathrm{n}}_1=0\dots \left(1\right) \mathrm{and} \mathrm{lx}+\mathrm{my}+{\mathrm{n}}_2=0 ...\left(2\right)$
$\begin{array}{l}\therefore {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=\mathrm{\lambda }\left(\mathrm{lx}+\mathrm{my}+{\mathrm{n}}_1\right)\left(\mathrm{lx}+\mathrm{my}+{\mathrm{n}}_2\right)\\ \end{array}$
Equating the like terms on both sides we get
${\mathrm{\lambda l}}^2=\mathrm{a} ...\left(3\right) 2\mathrm{\lambda lm}=2\mathrm{h} ...\left(4\right)$
${\mathrm{\lambda m}}^2=\mathrm{b} ...\left(5\right) \mathrm{\lambda l}\left({\mathrm{n}}_1+{\mathrm{n}}_2\right)=2\mathrm{g} ...\left(6\right)$
$\mathrm{\lambda m}\left({\mathrm{n}}_1+{\mathrm{n}}_2\right)=2\mathrm{f} ...\left(7\right) {\mathrm{\lambda n}}_1{\mathrm{n}}_2=\mathrm{c} ...\left(8\right)$
From (3) and (5), ${\mathrm{\lambda }}^2{\mathrm{l}}^2{\mathrm{m}}^2=\mathrm{ab} \mathrm{and} \mathrm{from} \left(4\right)$
${\mathrm{h}}^2={\mathrm{\lambda }}^2{\mathrm{l}}^2{\mathrm{m}}^2=\left({\mathrm{\lambda l}}^2\right)\left({\mathrm{\lambda m}}^2\right)=\mathrm{ab}$
Dividing (6) and (7), $\frac{\mathrm{l}}{\mathrm{m}}=\frac{\mathrm{g}}{\mathrm{f}}\Rightarrow \frac{{\mathrm{l}}^2}{{\mathrm{m}}^2}=\frac{{\mathrm{g}}^2}{{\mathrm{f}}^2}$
$\therefore \frac{\mathrm{a}}{\mathrm{b}}=\frac{{\mathrm{g}}^2}{{\mathrm{f}}^2}\Rightarrow {\mathrm{bg}}^2={\mathrm{af}}^2$
Distance between the parallel line (1) and (2) is
$\begin{array}{l}=\frac{\left|{\mathrm{n}}_1-{\mathrm{n}}_2\right|}{\sqrt{\left({\mathrm{l}}^2+{\mathrm{m}}^2\right)}}=\frac{\sqrt{{\left({\mathrm{n}}_1+{\mathrm{n}}_2\right)}^2-4{\mathrm{n}}_1{\mathrm{n}}_2}}{\sqrt{{\mathrm{l}}^2+{\mathrm{m}}^2}}\\ =\sqrt{\frac{{\left(\frac{2\mathrm{g}}{\mathrm{\lambda l}}\right)}^2-\frac{4\mathrm{c}}{\mathrm{\lambda }}}{\frac{\mathrm{a}}{\mathrm{\lambda }}+\frac{\mathrm{b}}{\mathrm{\lambda }}}}\text{ (or) }\sqrt{\frac{{\left(\frac{2\mathrm{f}}{\mathrm{\lambda m}}\right)}^2-\frac{4\mathrm{c}}{\mathrm{\lambda }}}{\frac{\mathrm{a}}{\mathrm{\lambda }}+\frac{\mathrm{b}}{\mathrm{\lambda }}}}\\ =\sqrt{\frac{4\left({\mathrm{g}}^2-{\mathrm{\lambda l}}^2\mathrm{c}\right)}{{\mathrm{l}}^2\mathrm{\lambda }(\mathrm{a}+\mathrm{b})}}\left(\text{ or }\right)\sqrt{\frac{4\left({\mathrm{f}}^2-{\mathrm{\lambda m}}^2\mathrm{c}\right)}{{\mathrm{m}}^2\mathrm{\lambda }(\mathrm{a}+\mathrm{b})}}\\ =2\sqrt{\left[\frac{{\mathrm{g}}^2-\mathrm{ac}}{\mathrm{a}(\mathrm{a}+\mathrm{b})}\right]}\text{ (or }\sqrt{\left[\frac{\left.{\mathrm{f}}^2-\mathrm{bc}\right]}{\mathrm{b}(\mathrm{a}+\mathrm{b})}\right]}\end{array}$