Question

Prove that the general equation
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
represents two parallel straight lines if
$h^2=ab$ and $b{g}^2=a{f}^2$
Prove also that the distance between them is
$\frac{2\sqrt{g^2-ac}}{a(a+b)}.$

Answer 1

${ax}^2+2hxy+b{y}^2+2gx+2fy+c=0{ax}^2+2(hy+g)x+(b{y}^2+2fy+c)=0$

Using quadratic formula

$x=\frac{-2(hy+g)\pm \sqrt{{4(hy+g)}^2-4a(b{y}^2+2fy+c)}}{2a}ax+hy+g=\pm \sqrt{{(hy+g)}^2-a(b{y}^2+2fy+c)}$

$ax+hy+g=\pm \sqrt{(h^2-ab)y^2+2(gh-af)y+(g^2-ac)}$ .... $\left(i\right)$

For parallel straight lines $R.H.S$ of $\left(i\right)$ must be independent of $y$

$\Rightarrow h^2-ab=0h^2=ab$

Also $gh-af=0$

$gh=af{g}^2{h}^2=a^2{f}^2{g}^{2}ab=a^2{f}^2\Rightarrow b{g}^2=a{f}^2$

Hence proved.

So the straight lines are given by $ax+hy+g=\pm \sqrt{(g^2-ac)}$

$ax+hy+g-\sqrt{(g^2-ac)}=0$

$ax+hy+g+\sqrt{(g^2-ac)}=0$

Distance between lines

$=\left|\frac{g-\sqrt{(g^2-ac)}-(g+\sqrt{(g^2-ac)})}{\sqrt{a^2+h^2}}\right|=\frac{2\sqrt{g^2-ac}}{\sqrt{a^2+h^2}}=\frac{2\sqrt{g^2-ac}}{\sqrt{a^2+ab}}=\frac{2\sqrt{g^2-ac}}{\sqrt{a(a+b)}}$