Question

Show that a homogeneous equation of degree two in $x$ and $y$, i.e., $a{x}^2+2hxy+h{y}^2=0$ represents a pair of lines passing through the origin if $h^2-ab\ge 0$.

Answer 1

Consider a homogeneous equation of degree two in $x$ and $y$, $a{x}^2+2hxy+h{y}^2=0...\left(1\right)$

In this equation at least one of the coefficients $a,borh$ is non zero.

We consider two cases.

Case I : If $b=0$, then the equation of lines are $x=0$ and $(ax+2hy)=0$.

These lines passes through the origin.

Case II : $b\ne 0$, Multiplying both the sides of equation $\left(1\right)$ by $b$, we get

$ab{x}^2+2hxyb+hb{y}^2=0$

$\Rightarrow hb{y}^2+2hbxy=-ab{x}^2$

To make L.H.S a complete square, we add $h^2{x}^2$ on both the sides.

$\Rightarrow hb{y}^2+2hbxy+h^2{x}^2=-ab{x}^2+h^2{x}^2$

$\Rightarrow (by+hx{)}^2=(h^2-ab)x^2$

$\Rightarrow (by+hx{)}^2={\left[\right(\sqrt{h^2-ab}\left)x\right]}^2$

$\Rightarrow (by+hx{)}^2-{\left[\right(\sqrt{h^2-ab}\left)x\right]}^2=0$

$\Rightarrow \left[\right(by+hx)-(\sqrt{h^2-ab}\left)x\right]\cdot \left[\right(by+hx)+(\sqrt{h^2-ab}\left)x\right]=0$

It is the joint equation of two lines $(by+hx)-\left(\sqrt{h^2-ab}\right)x=0$ and $(by+hx)+\left(\sqrt{h^2-ab}\right)x=0$

That is, $(h+\sqrt{h^2-ab})x+by=0$ and $(h-\sqrt{h^2-ab})x+by=0$

These lines passes through the origin.