Prove this equation represents a pair of straight lines. 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?
 A: Here are two ways to answer the question.


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*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.


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*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.
