# Prove that the homogeneous equation of second degree..

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.

Writing $ax^2+2b xy +c y^2$ for the quadratic form, the number of lines depends upon the eigenvalues of the matrix

$$\left(\begin{matrix} a & b \\ b & c \end{matrix} \right)$$

If the matrix is positive or negative definite there is only the origin when setting the form to zero (I assume we are in real space?). If zero is an eigenvalue there is one line only. If there is one positive, one negative eigenvalue you get two lines. Assume $a<>0$. You may rewrite the matrix in the form: $$\left(\begin{matrix} 1 & 0 \\ b/a & 1 \end{matrix} \right) \left(\begin{matrix} a & 0 \\ 0 & c-b^2/a \end{matrix} \right) \left(\begin{matrix} 1 & b/a \\ 0 & 1 \end{matrix} \right)$$ Here $c-b^2/a$ must be negative (to get two lines) and your quadratic form being zero gives: $$a (x+b/a\; y)^2 = (b^2/a - c)y^2$$ Taking $\pm$ square roots you get two lines. The case $c<>0$ is treated in a similar way and the case $a=c=0$ is easily solved.

Remark: there is probably a solution with a nicer symmetry in the constants.

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 $a,b,c.$

• Do you have any other simpler method, preferably shorter method? – pi-π Sep 7 '16 at 15:01
• Not really shorter. But one can show using the quadratic equation that, IF the discriminant $b^2-4ac>0,$ then the equation factors into two linear factors $dx+ey$ and $fx+gy,$ each of these being a line. I think these are two different lines, provided the discriminant is positive. – coffeemath Sep 7 '16 at 16:10