Prove that every homogeneous equation of second degree in $x$ and $y$ represents a pair of lines, each passing through the origin.
My Attempt:
Let $ax^2+2hxy+by^2=0$ be a homogeneous equation of second degree in $x$ and $y$.
We can write this equation as
$$by^2+2hxy+ax^2=0$$
Dividing both sides by $x^2$,
$$b\frac {y^2}{x^2} + 2h\frac {y}{x} +a=0$$.
Now, what should I do to continue.
Please help.
Best Answer
You have divided by $x^2,$ but no loss there since if $x^2=0$ then $x=0$ and that leads to $y=0$ provided $b \neq 0.$ In that case the point $(0,0)$ is on it.
If it happens that $b=0,$ then it factors as $x(a+2hy)=0,$ and setting each factor to zero gives a line through the origin as desired.
Now assume $x \neq 0$ so your steps so far are OK. Then your final equation is a quadratic in the quantity $u=y/x$ (unless $b=0$ already dealt with). So the next step would be to solve this quadratic for $u.$ If you can show you always get two solutions, then that leads back to two lines on putting $y/x$ equal to the two solutions of the quadratic.
I haven't checked what happens if there's a double root, or imaginary roots.
Added note: the equation $y^2=0$ is homogeneous and represents only one line in the plane, namely the $x$ axis. As another example, only the point $(0,0)$ satisfies $x^2+xy+y^2=0$ so in that case the equation represents only a single point (the origin) which would likely not qualify as "two lines." So to make sure one gets two lines extra condition(s) are needed on $a,b,c.$