Prove this equation represents a pair of straight lines.

Question

I am to prove that $(ab-h^2)(ax^2+2hxy+by^2+2gx+2fy)+af^2+bg^2-2fgh=0$ represents a pair of straight lines. I am aware of the condition that is represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$ in the general equation for 2nd degree, but that would turn out to be very cumbersome. So, is there any shortcut to prove this?

Answer 1

Here are two ways to answer the question.

• 1) A matrix oriented arguing:

It is clear that the following condition must hold:

$$\tag{0}m:=a b - h^2 \neq 0$$

(otherwise, the LHS of the given equation would become a constant).

The equation of the conical section can thus be divided by $m$, and therefore be written

$$\tag{1}U:=ax^2+2hxy+by^2+2gx+2fy+c=0 \ \ \text{where} \ \ c:=\frac{a f^2 + b g^2 - 2f g h}{a b - h^2}$$

or, in a matrix form:

$U=X^TMX=0$ with $X^T=(x \ y \ 1)$ and:

$$\tag{2} M=\pmatrix{a& h& g\\h& b& f\\g& f& c}.$$

Expanding the determinant of $M$ along its third column gives $det(M)=0$.

Therefore $rank(M)\leq 2$.

But condition (0) expresses the fact that the upper left minor of $M$ is nonzero. Thus $rank(M)\geq 2$. Finally:

$$\tag{3} rank(M)=2.$$

Thus the conical section is decomposable into 2 straight lines (see for example (Decomposition of a degenerate conic)), but these straight lines might have complex coefficients!

It appears that a condition of reality is

$$\tag{3}h^2 > ab$$

(were you aware of it?) as we are going to see it in the second way.

• 2) An explicit way, by decomposition of $U$.

Under condition (3), (1) can be written

$$\tag{4}U=\left(s x + \frac{h y + g}{s}\right)^2-\left(t y + \frac{u}{t}\right)^2=0$$

by setting $s=\sqrt{a}, t=\dfrac{\sqrt{h^2 - a b}}{s}, u=\dfrac{g h - a f}{a}.$

Being a difference of two squares, (4) can be decomposed into two pairs of first degree equations which are the equations of the straight lines.